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LIBRARY OF 
WELLESLEY ‘COLLECE 


PURCHASED FROM 
LIBRARY FUNDS 


THe WORKS 


OF 


ARCHIME Dis. 


London: C. J. CLAY anv SONS, 
CAMBRIDGE UNIVERSITY PRESS WAREHOUSE, 
AVE MARIA LANE. 

Glasgow: 263, ARGYLE STREET. 


Leipsig: F. A. BROCKHAUS. 
fet Work: THE MACMILLAN COMPANY. 


THE WORKS 


OF 


ΠΟ LH lM BEDE 5 


2 


EDITED IN MODERN NOTATION 


WITH INTRODUCTORY CHAPTERS 


BY 


rE 1. HEATH sesh, 


SOMETIME FELLOW OF TRINITY COLLEGE, CAMBRIDGE. 


CAMBRIDGE: 
AT THE UNIVERSITY PRESS. 


1897 


[All Rights reserved.] 


LO |e 


Cambridge : 
PRINTED BY J. AND C. F. CLAY, 
AT THE UNIVERSITY PRESS. 


Ne δ 


ΤΗΝ έν .., 


PREFACE. 


HIS book is intended to form a companion volume to my 
edition of the treatise of Apollonius on Conic Sections 
lately published. If it was worth while to attempt to make the 
work of “the great geometer” accessible to the mathematician 
of to-day who might not be able, in consequence of its length 
and of its form, either to read it in the original Greek or in a 
Latin translation, or, having read it, to master it and grasp the 
whole scheme of the treatise, I feel that I owe even less of an 
apology for offering to the public a reproduction, on the same 
lines, of the extant works of perhaps the greatest mathematical 
genius that the world has ever seen. 

Michel Chasles has drawn an instructive distinction between 
the predominant features of the geometry of Archimedes and 
of the geometry which we find so highly developed in Apollo- 
nius. Their works may be regarded, says Chasles, as the origin 
and basis of two great inquiries which seem to share between 
them the domain of geometry. Apollonius is concerned with 
the Geometry of Forms and Situations, while in Archimedes 
we find the Geometry of Measurements dealing with the quad- 
rature of curvilinear plane figures and with the quadrature 
and cubature of curved surfaces, investigations which “gave 
birth to the calculus of the infinite conceived and brought 
to perfection successively by Kepler, Cavalieri, Fermat, Leibniz, 
and Newton.” But whether Archimedes is viewed as the 
man who, with the limited means at his disposal, nevertheless 
succeeded in performing what are really integrations for the 
purpose of finding the area of a parabolic segment and a 


Vi PREFACE. 


spiral, the surface and volume of a sphere and a segment 
of a sphere, and the volume of any segments of the solids 
of revolution of the second degree, whether he is seen finding 
the centre of gravity of a parabolic segment, calculating 
arithmetical approximations to the value of 7, inventing a 
system for expressing in words any number up to that which 
we should write down with 1 followed by 80,000 billion 
ciphers, or inventing the whole science of hydrostatics and at 
the same time carrying it so far as to give a most complete 
investigation of the positions of rest and stability of a right 
segment of a paraboloid of revolution floating in a fluid, the 
intelligent reader cannot fail to be struck by the remarkable 
range of subjects and the mastery of treatment. And if these 
are such as to create genuine enthusiasm in the student of 
Archimedes, the style and method are no less irresistibly 
attractive. One feature which will probably most impress the 
mathematician accustomed to the rapidity and directness secured 
by the generality of modern methods is the deliberation with 
which Archimedes approaches the solution of any one of his 
main problems. Yet this very characteristic, with its incidental 
effects, is calculated to excite the more admiration because the 
method suggests the tactics of some great strategist who 
foresees everything, eliminates everything not immediately 
conducive to the execution of his plan, masters every position 
in its order, and then suddenly (when the very elaboration of 
the scheme has almost obscured, in the mind of the spectator, 
its ultimate object) strikes the final blow. Thus we read in 
Archimedes proposition after proposition the bearing of which is 
not immediately obvious but which we find infallibly used later 
on; and we are led on by such easy stages that the difficulty of 
the original problem, as presented at the outset, is scarcely 
appreciated. As Plutarch says, “it is not possible to find in 
geometry more difficult and troublesome questions, or more 
simple and lucid explanations.” But it is decidedly a rhetorical 
exaggeration when Plutarch goes on to say that we are deceived 


PREFACE. Vil 


by the easiness of the successive steps into the belief that anyone 
could have discovered them for himself. On the contrary, the 
studied simplicity and the perfect finish of the treatises involve 
at the same time an element of mystery. Though each step 
depends upon the preceding ones, we are left in the dark as to 
how they were suggested to Archimedes. There is, in fact, 
much truth in a remark of Wallis to the effect that he seems 
“as it were of set purpose to have covered up the traces of his 
investigation as if he had grudged posterity the secret of his 
method of inquiry while he wished to extort from them assent 
to his results.” Wallis adds with equal reason that not only 
Archimedes but nearly all the ancients so hid away from 
posterity their method of Analysis (though it is certain that 
they had one) that more modern mathematicians found it easier 
to invent a new Analysis than to seek out the old. This is no 
doubt the reason why Archimedes and other Greek geometers 
have received so little attention during the present century and 
why Archimedes is for the most part only vaguely remembered 
as the inventor of a screw, while even mathematicians scarcely 
know him except as the discoverer of the principle in hydro- 
statics which bears his name. It is only of recent years that 
we have had a satisfactory edition of the Greek text, that of 
Heiberg brought out in 1880-1, and I know of no complete 
translation since the German one of Nizze, published in 1824, 
which is now out of print and so rare that I had some difficulty 
in procuring a copy. 

The plan of this work is then the same as that which I 
followed in editing the Conics of Apollonius. In this case, 
however, there has been less need as well as less opportunity for 
compression, and it has been possible to retain the numbering 
of the propositions and to enunciate them in a manner more 
nearly approaching the original without thereby making the 
enunciations obscure. Moreover, the subject matter is not so 
complicated as to necessitate absolute uniformity in the notation 
used (which is the only means whereby Apollonius can be made 


Vill PREFACE. 


even tolerably readable), though I have tried to secure as much 
uniformity as was fairly possible. My main object has been to 
present a perfectly faithful reproduction of the treatises as they 
have come down to us, neither adding anything nor leaving out 
anything essential or important. The notes are for the most 
part intended to throw light on particular points in the text or 
to supply proofs of propositions assumed by Archimedes as 
known; sometimes I have thought it right to insert within 
square brackets after certain propositions, and in the same type, 
notes designed to bring out the exact significance of those 
propositions, in cases where to place such notes in the Intro- 
duction or at the bottom of the page might lead to their being 
overlooked. 

Much of the Introduction is, as will be seen, historical; the 
rest is devoted partly to giving a more general view of certain 
methods employed by Archimedes and of their mathematical 
significance than would be possible in notes to separate propo- 
sitions, and partly to the discussion of certain questions arising 
out of the subject matter upon which we have no positive 
historical data to guide us. In these latter cases, where it is 
necessary to put forward hypotheses for the purpose of explaining 
obscure points, I have been careful to call attention to their 
speculative character, though I have given the historical evidence 
where such can be quoted in support of a particular hypothesis, 
my object being to place side by side the authentic information 
which we possess and the inferences which have been or may 
be drawn from it, in order that the reader may be in a position 
to judge for himself how far he can accept the latter as probable. 
Perhaps I may be thought to owe an apology for the length of 
one chapter on the so-called νεύσεις, or inclinationes, which goes 
somewhat beyond what is necessary for the elucidation of 
Archimedes; but the subject is interesting, and I thought it 
well to make my account of it as complete as possible in 
order to round off, as it were, my studies in Apollonius and 
Archimedes. 


PREFACE. 1x 


I have had one disappointment in preparing this book for 
the press. I was particularly anxious to place on or opposite 
the title-page a portrait of Archimedes, and I was encouraged 
in this idea by the fact that the title-page of Torelli’s edition 
bears a representation in medallion form on which are endorsed 
the words ‘Archimedis effigies marmorea in vetert anaglypho 
Romae asservato. Caution was however suggested when I 
found two more portraits wholly unlike this but still claiming to 
represent Archimedes, one of them appearing at the beginning 
of Peyrard’s French translation of 1807, and the other in 
Gronovius’ Thesaurus Graecarum Antiquitatum ; and I thought 
it well to inquire further into the matter. I am now informed 
by Dr A. S. Murray of the British Museum that there does 
not appear to be any authority for any one of the three, and 
that writers on iconography apparently do not recognise an 
Archimedes among existing portraits. I was, therefore, re- 
luctantly obliged to give up my idea. 

The proof sheets have, as on the former occasion, been read 
over by my brother, Dr R. 5. Heath, Principal of Mason College, 
Birmingham ; and I desire to take this opportunity of thanking 
him for undertaking what might well have seemed, to any one 
less genuinely interested in Greek geometry, a thankless task. 


Τ. 1. BATH. 


March, 1897. 


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LIST OF THE PRINCIPAL WORKS CONSULTED. 


JOSEPH TORELLI, Archimedis quae supersunt omnia cum LHutocii Asca- 
lonitae commentariis. (Oxford, 1792.) 

Ernst ΝΊΖΖΕ, Archimedes von Syrakus vorhandene Werke aus dem 
griechischen tibersetzt und mit erliuternden und kritischen Anmerk- 
ungen begleitet. (Stralsund, 1824.) 

J. L. Herpere, Archimedis opera omnia cum commentariis Eutociv. 
(Leipzig, 1880-1.) 

J. Τὸ HerBereG, Quaestiones Archimedeae. (Copenhagen, 1879.) 

F. Huutscn, Article Archimedes in Pauly-Wissowa’s Real-Encyclopidie der 
classischen Altertumswissenschaften. (Edition of 1895, τι. 1, pp. 
507-539.) 


C. A. BRETSCHNEIDER, Die Geometrie und die Geometer vor Euklides. 
(Leipzig, 1870.) 

M. Cantor, Vorlesungen iiber Geschichte der Mathematik, Band 1, zweite 
Auflage. (Leipzig, 1894.) 

G. Frrep.etn, Procli Diadochi in primum Euclidis elementorum librum 
commentarii. (Leipzig, 1873.) 

JAMES Gow, A short history of Greek Mathematics. (Cambridge, 1884.) 

StzemunD GUtnrHerR, Abriss der Geschichte der Mathematik und der 
Naturwissenschaften im Altertum in Iwan von Miiller’s Handbuch der 
klassischen Altertumswissenschaft, v. 1. 

HERMANN HANKEL, Zur Geschichte der Mathematik in Alterthum und 
Mittelalter. (Leipzig, 1874.) 

J. Τὸ Herpere, Litterargeschichtliche Studien iiber Euklid. (Leipzig, 
1882.) 

J. L. Herpere, Luclidis elementa. (Leipzig, 1883-8.) 

F. Huxtscn, Article Arithmetica in Pauly-Wissowa’s Real-Encyclopiidie, 
i. 1, pp. 1066-1116. 


ΧΙ LIST OF PRINCIPAL WORKS CONSULTED. 


F. Houurscw, Heronis Alexandrini geometricorum et stereometricorum 
reliquiae. (Berlin, 1864.) 

F. Huurscw, Pappi Alexandrini collectionis quae supersunt. (Berlin, 
1876-8.) 

Grvo Loria, Jl periodo aureo della geometria greca. (Modena, 1895.) 


Maximitien Mariz, Histoire des sciences mathématiques et physiques, 
Tome I. (Paris, 1883.) 

J. H. T. Miuuer, Bertriige zur Terminologie der griechischen Mathematiker. 
(Leipzig, 1860.) 

G. H. F. Nessetmann, Die Algebra der Griechen. (Berlin, 1842.) 

F. SusEMIHL, Geschichte der griechischen Litteratur in der Alexandrinerzett, 
Band 1. (Leipzig, 1891.) 

P. Tannery, La Géométrie grecque, Premiére partie, Histoire générale de la 
Géometrie élémentaire. (Paris, 1887.) 

H. G. ZeutHen, Die Lehre von den Kegelschnitten im Altertum. (Copen- 
hagen, 1886.) 

H. G. ZeurHen, Geschichte der Mathematik im Altertum und Mittelalter. 
(Copenhagen, 1896.) 


CHAPTER I. 


CHAPTER II. 


CuHapter ITI. 


ΕΠ 
§ 2. 


CHAPTER IV. 


§ 1. 
ὃ 2. 
ὅτι 
§ 4. 
ὃ 5. 
ὃ 6. 


§ 7. 
§ 8. 


CONTENTS. 
INTRODUCTION. 


ARCHIMEDES 


MANUSCRIPTS AND PRINCIPAL EDITIONS—ORDER 
OF COMPOSITION—DIALECT—LOST WORKS 


RELATION OF ARCHIMEDES TO HIS PREDECESSORS 

Use of traditional geometrical methods 

Earlier discoveries affecting quadrature and 
cubature . 

Conic Sections 

Surfaces of the second Slee : 

Two mean proportionals in continued Sonor: 
tion . 


ARITHMETIC IN ARCHIMEDES 

Greek numeral system 

Addition and subtraction . 

Multiplication 

Division : 

Extraction of fib square root : 

Early investigations of surds or incommensu- 
rables 

Archimedes’ sprronmations to V3 

Archimedes’ approximations to the square roots 
of large numbers which are not complete 
squares 

Note on alternative πον Ati spent © 
the approximations to ./3 


PAGE 
XV 


XXill 


XXX1X 
xl 


xlvii 
li 

liv 
lxvii 
xviii 
xix 
ΙΧΧῚ 
ΙΧΧῚΪ 
lxxili 
|xxiv 


Ixxvil 
Ixxx 


lxxxiv 


ΧΟ 


X1V 


CHAPTER V. 


CHapter VI. 


CuHapter VII. 


CONTENTS. 


ON THE PROBLEMS KNOWN AS NEYSEIS 

Nevoes referred to by Archimedes 

Mechanical constructions: the conchoid of Nico- 
medes : 

Pappus’ solution of the νεῦσις reed Ὧ in 
Props. 8, 9 On Spirals . 

The problem of the two mean pear anipeate 

The trisection of an angle 

On certain plane vevoes 


CUBIC EQUATIONS 


ANTICIPATIONS BY ARCHIMEDES OF THE INTE- 
GRAL CALCULUS 


CHapteR VIII. THE TERMINOLOGY OF ARCHIMEDES . 


THE WORKS OF ARCHIMEDES. 


ON THE SPHERE AND CYLINDER, BOOK I. . 


” 


MEASUREME 
ON CONOIDS 
ON SPIRALS 


” ” ” BOOK II. 
NT OF A CIRCLE 
AND SPHEROIDS 


ON THE EQUILIBRIUM OF PLANES, BOOK I. 


9? 


BOOK II. 


” 37 ” 


THE SAND-RECKONER 
QUADRATURE OF THE PARABOLA 
ON FLOATING BODIES, BOOK I. 


” ” 


5: BOOK II. 


BOOK OF LEMMAS. 
THE CATTLE-PROBLEM 


PAGE 


CV 


151 
189 
203 
221 
233 
253 
263 
301 
319 


INTRODUCTION. 


CHAPTER I. 
ARCHIMEDES, 


A ΠΡῈ of Archimedes was written by one Heracleides*, but 
this biography has not survived, and such particulars as are known 
have to be collected from many various sources}. According to 
Tzetzest he died at the age of 75, and, as he perished in the sack 
of Syracuse (B.c. 212), it follows that he was probably born about 
287 B.c. He was the son of Pheidias the astronomer§, and was 
on intimate terms with, if not related to, king Hieron and his 


* Kutocius mentions this work in his commentary on Archimedes’ Measwre- 
ment of the circle, ὥς φησιν ‘Hpaxdeldns ἐν τῷ ᾿Αρχιμήδους βίῳ. He alludes te it 
again in his commentary on Apollonius’ Conics (ed. Heiberg, Vol. τι. p. 168), 
where, however, the name is wrongly given as Ἡράκλειος. This Heracleides is 
perhaps the same as the Heracleides mentioned by Archimedes himself in the 
preface to his book On Spirals. 

+ An exhaustive collection of the materials is given in Heiberg’s Quaestiones 
Archimedeae (1879). The preface to Torelli’s edition also gives the main points, 
and the same work (pp. 363—370) quotes at length most of the original 
references to the mechanical inventions of Archimedes. Further, the article 
Archimedes (by Hultsch) in Pauly-Wissowa’s Real-Encyclopidie der classischen 
Altertumswissenschaften gives an entirely admirable summary of all the available 
information. See also Susemihl’s Geschichte der griechischen Litteratur in der 
Alexandrinerzeit, τ. pp. 723—733. 

+ Tzetzes, Chiliad., 11. 35, 105. 

§ Pheidias is mentioned in the Sand-reckoner of Archimedes, τῶν προτέρων 
ἀστρολόγων Hvddeov...Pecdla δὲ τοῦ ἁμοῦ πατρὸς (the last words being the correction 
of Blass for τοῦ ᾿Ακούπατρος, the reading of the text). Cf. Schol. Clark. in 
Gregor. Nazianz. Or. 34, p. 355a Morel. Φειδίας τὸ μὲν γένος ἢν Συρακόσιος 
ἀστρολόγος ὁ ᾿Αρχιμήδους πατήρ. 


Xvl INTRODUCTION. 


son Gelon. It appears from a passage of Diodorus* that he spent 
a considerable time at Alexandria, where it may be inferred that 
he studied with the successors of Euclid. It may have been at 
Alexandria that he made the acquaintance of Conon of Samos 
(for whom he had the highest regard both as a mathematician 
and as a personal friend) and of Eratosthenes. To the former 
he was in the habit of communicating his discoveries before their 
publication, and it is to the latter that the famous Cattle-problem 
purports to have been sent. Another friend, to whom he dedicated 
several of his works, was Dositheus of Pelusium, a pupil of Conon, 
presumably at Alexandria though at a date subsequent to Archi- 
medes’ sojourn there. 

After his return to Syracuse he lived a life entirely devoted 
to mathematical research. Incidentally he made himself famous 
by a variety of ingenious mechanical inventions. These things 
were however merely the ‘diversions of geometry at play t,” and 
he attached no importance to them. In the words of Plutarch, “he 
possessed so high a spirit, so profound a soul, and such treasures 
of scientific knowledge that, though these inventions had obtained 
for him the renown of more than human sagacity, he yet would 
not deign to leave behind him any written work on such subjects, 
but, regarding as ignoble and sordid the business of mechanics 
and every sort of art which is directed to use and profit, he placed 
his whole ambition in those speculations in whose beauty and 
subtlety there is no admixture of the common needs of life{.” In 
fact he wrote only one such mechanical book, On Sphere-making§, 
to which allusion will be made later. : 

Some of his mechanical inventions were used with great effect 
against the Romans during the siege of Syracuse. Thus he contrived 


* Diodorus v. 37, 3, ods [τοὺς κοχλίας] ᾿Αρχιμήδης ὁ Συρακόσιος εὗρεν, ὅτε 
παρέβαλεν εἰς Αἴγυπτον. 

+ Plutarch, Marcellus, 14. 

ac ἡδιτα. 17: 

8 Pappus vu. p. 1026 (ed. Hultsch). Κάρπος δὲ πού φησιν ὁ ᾿Αντιοχεὺς 
᾿Αρχιμήδη τὸν Συρακόσιον ἕν μόνον βιβλίον συντεταχέναι μηχανικὸν τὸ κατὰ τὴν 
σφαιροποιΐαν, τῶν δὲ ἄλλων οὐδὲν ἠξιωκέναι συντάξαι. καίτοι παρὰ τοῖς πολλοῖς ἐπὶ 
μηχανικῇ δοξασθεὶς καὶ μεγαλοφυής τις γενόμενος ὁ θαυμαστὸς ἐκεῖνος, ὥστε διαμεῖναι 
παρὰ πᾶσιν ἀνθρώποις ὑπερβαλλόντως ὑμνούμενος, τῶν τε προηγουμένων γεωμετρικῆς 
καὶ ἀριθμητικῆς ἐχομένων θεωρίας τὰ βραχύτατα δοκοῦντα εἶναι σπουδαίως συνέγραφεν" 
ὃς φαίνεται τὰς εἰρημένας ἐπιστήμας οὕτως ἀγαπήσας ὡς μηδὲν ἔξωθεν ὑπομένειν 


αὐταῖς ἐπεισάγειν. 


ARCHIMEDES. xvil 


catapults so ingeniously constructed as to be equally serviceable 
at long or short ranges, machines for discharging showers of 
missiles through holes made in the walls, and others consisting 
of long moveable poles projecting beyond the walls which either 
dropped heavy weights upon the enemy’s ships, or grappled the 
prows by means of an iron hand or a beak like that of a crane, 
then lifted them into the air and let them fall again*. Marcellus 
is said to have derided his own engineers and artificers with the 
words, “Shall we not make an end of fighting against this geo- 
metrical Briareus who, sitting at ease by the sea, plays pitch and 
toss with our ships to our confusion, and by the multitude of 
missiles that he hurls at us outdoes the hundred-handed giants of 
mythology!+”; but the exhortation had no effect, the Romans being 
in such abject terror that “if they did but see a piece of rope 
or wood projecting above the wall, they would cry ‘there it is 
again,’ declaring that Archimedes was setting some engine in motion 
against them, and would turn their backs and run away, insomuch 
that Marcellus desisted from all conflicts and assaults, putting all 
his hope in a long sieget.” 

If we are rightly informed, Archimedes died, as he had lived, 
absorbed in mathematical contemplation. The accounts of the 
exact circumstances of his death differ in some details. Thus 
Livy says simply that, amid the scenes of confusion that followed 
the capture of Syracuse, he was found intent on some figures which 
he had drawn in the dust, and was killed by a soldier who did 
not know who he was§. Plutarch gives more than one version in 
the following passage. ‘‘ Marcellus was most of all afflicted at 
the death of Archimedes ; for, as fate would have it, he was intent 
on working out some problem with a diagram and, having fixed 
his mind and his eyes alike on his investigation, he never noticed 
the incursion of the Romans nor the capture of the city. And 
when a soldier came up to him suddenly and bade him follow to 


* Polybius, Hist. vin. 7—8; Livy xxiv. 34; Plutarch, Marcellus, 15—17. 

+ Plutarch, Marcellus, 17. 

+ ibid. 

§ Livy xxv. 31. Cum multa irae, multa auaritiae foeda exempla ederentur, 
Archimedem memoriae proditum est in tanto tumultu, quantum pauor captae 
urbis in discursu diripientium militum ciere poterat, intentum formis, quas in 
puluere descripserat, ab ignaro milite quis esset interfectum ; aegre id Marcellum 
tulisse sepulturaeque curam habitam, et propinquis etiam inquisitis honori 
praesidioque nomen ac memoriam eius fuisse. 


H. A. δ 


XVlll INTRODUCTION. 


Marcellus, he refused to do so until he had worked out his problem 
to a demonstration; whereat the soldier was so enraged that he 
drew his sword and slew him, Others say that the Roman ran 
up to him with a drawn sword offering to kill him; and, when 
Archimedes saw him, he begged him earnestly to wait a short time 
in order that he might not leave his problem incomplete and 
unsolved, but the other took no notice and killed him. Again 
there is a third account to the effect that, as he was carrying to 
Marcellus some of his mathematical instruments, sundials, spheres, 
and angles adjusted to the apparent size of the sun to the sight, some 
soldiers met him and, being under the impression that he carried 
gold in the vessel, slew him*.” The most picturesque version of the 
story is perhaps that which represents him as saying to a Roman 
soldier who came too close, “Stand away, fellow, from my diagram,” 
whereat the man was so enraged that he killed himy+. The addition 
made to this story by Zonaras, representing him as saying παρὰ 
κεφαλὰν καὶ μὴ παρὰ γραμμάν, while it no doubt recalls the second 
version given by Plutarch, is perhaps the most far-fetched of the 
touches put to the picture by later hands. 

Archimedes is said to have requested his friends and relatives 
to place upon his tomb a representation of a cylinder circumscribing 
a sphere within it, together with an inscription giving the ratio 
which the cylinder bears to the sphere}; from which we may 
infer that he himself regarded the discovery of this ratio [On the 
Sphere and Cylinder, 1. 33, 34] as his greatest achievement. Cicero, 
when quaestor in Sicily, found the tomb in a neglected state and 
restored itS. 

Beyond the above particulars of the life of Archimedes, we 
have nothing left except a number of stories, which, though perhaps 
not literally accurate, yet help us to a conception of the personality 
of the most original mathematician of antiquity which we would 
not willingly have altered. Thus, in illustration of his entire 
preoccupation by his abstract studies, we are told that he would 
forget all about his food and such necessities of life, and would 
be drawing geometrical figures in the ashes of the fire, or, when 


Plutarch, Marcellus, 19. 

Tzetzes, Chil. 11. 85, 1385; Zonaras Ix. 5. 
Plutarch, Marcellus, 17 ad jin. 

Cicero, Tusc. vy. 64 sq, 


ὧν tt + * 


ARCHIMEDES. ΧΙΧ 


anointing himself, in the oil on his body*. Of the same kind is 
the well-known story that, when he discovered in a bath the 
solution of the question referred to him by Hieron as to whether 
a certain crown supposed to have been made of gold did not in 
reality contain a certain proportion of silver, he ran naked through 
the street to his home shouting εὕρηκα, εὕρηκα . 

According to Pappust it was in connexion with his discovery 
of the solution of the problem Zo move a gwen weight by a given 
force that Archimedes uttered the famous saying, “Give me a 
place to stand on, and I can move the earth (δός μοι ποῦ στῶ καὶ 
κινῶ τὴν γῆν). Plutarch represents him as declaring to Hieron 
that any given weight could be moved by a given force, and 
boasting, in reliance on the cogency of his demonstration, that, if 
he were given another earth, he would cross over to it and move 
this one. ‘And when Hieron was struck with amazement and asked 
him to reduce the problem to practice and to give an illustration 
of some great weight moved by a small force, he fixed upon a ship 
of burden with three masts from the king’s arsenal which had 
only been drawn up with great labour and many men; and loading 
her with many passengers and a full freight, sitting himself the 
while far off, with no great endeavour but only holding the end 
of a compound pulley (πολύσπαστος) quietly in his hand and pulling 
at it, he drew the ship along smoothly and safely as if she were 
moving through the sea$.” According to Proclus the ship was one 
which Hieron had had made to send to king Ptolemy, and, when all 
the Syracusans with their combined strength were unable to launch 
it, Archimedes contrived a mechanical device which enabled Hieron 
to move it by himself, insomuch that the latter declared that 
“from that day forth Archimedes was to be believed in every- 
thing that he might say|.” While however it is thus established 
that Archimedes invented some mechanical contrivance for moving 
a large ship and thus gave a practical illustration of his thesis, 
it is not certain whether the machine used was simply a compound 


* Plutarch, Marcellus, 17. 

+ Vitruvius, Architect. 1x. 3. For an explanation of the manner in which 
Archimedes probably solved this problem, see the note following On jloating 
bodies, τ. 7 (p. 259 sq.). 

+ Pappus vir. p. 1060. 

§ Plutarch, Marcellus, 14. 

|| Proclus, Comm. on Eucl. 1., Ὁ. 63 (ed. Friedlein). 


62 


XX INTRODUCTION. 


pulley (πολύσπαστος) as stated by Plutarch; for Athenaeus*, in 
describing the same incident, says that a heliw was used. This 
term must be supposed to refer to a machine similar to the κοχλίας 
described by Pappus, in which a cog-wheel with oblique teeth 
moves on a cylindrical helix turned by a handley. Pappus, how- 
ever, describes it in connexion with the βαρουλκός of Heron, and, 
while he distinctly refers to Heron as his authority, he gives no 
hint that Archimedes invented either the βαρουλκός or the par- 
ticular κοχλίας ; on the other hand, the πολύσπαστος is mentioned 
by Galen 1, and the τρίσπαστος (triple pulley) by Oribasius§, as one 
of the inventions of Archimedes, the τρίσπαστος being so called 
either from its having three wheels (Vitruvius) or three ropes 
(Oribasius). Nevertheless, it may well be that though the ship 
could easily be kept in motion, when once started, by the τρί: 
σπαστος or πολύσπαστος, Archimedes was obliged to use an appliance 
similar to the κοχλίας to give the first impulse. 

The name of yet another instrument appears in connexion with 
the phrase about moving the earth. Tzetzes’ version is, ‘‘ Give 
me a place to stand on (za Bw), and I will move the whole earth 
with a χαριστίων ||”; but,.as in another passage] he uses the word 
τρίσπαστος, it may be assumed that the two words represented one 
and the same thing**. 

It will be convenient to mention in this place the other 
mechanical inventions of Archimedes. The best known is the 


* Athenaeus v. 207 a—b, κατασκευάσας yap ἕλικα TO τηλικοῦτον σκάφος εἰς THY 
θάλασσαν κατήγαγε" πρῶτος δ᾽ ᾿Αρχιμήδης εὗρε τὴν τῆς ἕλικος κατασκευήν. ΤῸ the 
same effect is the statement of Eustathius ad Il. 111. p. 114 (ed. Stallb.) λέγεται 
δὲ ἕλιξ Kal τι μηχανῆς εἶδος, ὃ πρῶτος εὑρὼν ὁ ᾿Αρχιμήδης εὐδοκίμησέ, φασι, dc αὐτοῦ, 

+ Pappus virt. pp. 1066, 1108 sq. 

+ Galen, in Hippocr. De artic., 1v. 47 (=xvut. p. 747, ed. Kiihn). 

§ Oribasius, Coll. med., xu1x. 22 (iv. p. 407, ed. Bussemaker), ᾿Απελλίδους ἢ 
᾿Αρχιμήδους τρίσπαστον, described in the same passage as having been invented 
πρὸς Tas τῶν πλοίων καθολκάς. 

|| Tzetzes, Chil. τι. 180. 

q Ibid., ut. 61, ὁ γῆν ἀνασπῶν μηχανῇ τῇ τρισπάστῳ βοῶν' ὅπα Bw καὶ σαλεύσω 
τὴν χθόνα. 

** Heiberg compares Simplicius, Comm. in Aristot. Phys. (ed. Diels, p. 1110, 
1. 2), ταύτῃ δὲ τῇ ἀναλογίᾳ τοῦ κινοῦντος καὶ τοῦ κινουμένου Kal τοῦ διαστήματος 
τὸ σταθμιστικὸν ὄργανον τὸν καλούμενον χαριστίωνα συστήσας ὁ ᾿Αρχιμήδης ὡς 
μέχρι παντὸς τῆς ἀναλογίας προχωρούσης ἐκόμπασεν ἐκεῖνο τὸ πᾶ BQ καὶ κινώ τὰν 
γᾶν. 


ARCHIMEDES. ΧΧῚ 


water-screw* (also called κοχλίας) which was apparently invented 
by him in Egypt, for the purpose of irrigating fields. It was 
also used for pumping water out of mines or from the hold of 
ships. 

Another invention was that of a sphere constructed so as to 
imitate the motions of the sun, the moon, and the five planets 
in the heavens. Cicero actually saw this contrivance and gives a 
description of itt, stating that it represented the periods of the 
moon and the apparent motion of the sun with such accuracy that 
it would even (over a short period) show the eclipses of the sun 
and moon. Hultsch conjectures that it was moved by water {. 
We know, as above stated, from Pappus that Archimedes wrote 
a book on the construction of such a sphere (περὶ σφαιροποιἴας), 
and Pappus speaks in one place of “those who understand the 
making of spheres and produce a model of the heavens by means 
of the regular circular motion of water.” In any case it is certain 
that Archimedes was much occupied with astronomy. Livy calls 
him “unicus spectator caeli siderumque.” Hipparchus says§, 
“From these observations it is clear that the differences in the 
years are altogether small, but, as to the solstices, I almost 
think (οὐκ ἀπελπίζω) that both I and Archimedes have erred to 
the extent of a quarter of a day both in the observation and in the 
deduction therefrom.” It appears therefore that Archimedes had 
considered the question of the length of the year, as Ammianus 
also states|). Macrobius says that he discovered the distances of 
the planets{/. Archimedes himself describes in the Sand-reckoner 
the apparatus by which he measured the apparent diameter of the 
sun, or the angle subtended by it at the eye. 

The story that he set the Roman ships on fire by an arrange- 
ment of burning-glasses or concave mirrors is not found in any 


* Diodorus 1. 34, v. 837; Vitruvius x. 16 (11); Philo m1. p. 330 (ed. Pfeiffer) ; 
Strabo xvi. p. 807; Athenaeus v. 208 f. 

+ Cicero, De rep., τ. 21-22; Tusc., 1.63; De nat. deor., τι. 88. Cf. Ovid, 
Fasti, νι. 277; Lactantius, Instit., τι. 5, 18; Martianus Capella, 1. 212, νι. 
583 sq.; Claudian, Epigr. 18; Sextus Empiricus, p. 416 (ed. Bekker). 

+ Zeitschrift f. Math. wu. Physik (hist. litt. Abth.), xx. (1877), 106 sq. 

8 Ptolemy, σύνταξις, τ. p. 153. 

|| Ammianus Marcell., xxvr. i. 8. 

4 Macrobius, in Somn. Scip., 11. 3. 


XX1l INTRODUCTION. 


authority earlier than Lucian*; and the so-called loculus Archi- 
medius, which was a sort of puzzle made of 14 pieces of ivory of 
different shapes cut out of a square, cannot be supposed to be his 
invention, the explanation of the name being perhaps that it was 
only a method of expressing that the puzzle was cleverly made, 
in the same way as the πρόβλημα ᾿Αρχιμήδειον came to be simply 
a proverbial expression for something very difficult Ἷ. 


* The same story is told of Proclus in Zonaras xtv. 3. For the other 
references on the subject see Heiberg’s Quaestiones Archimedeae, pp. 39-41. 
+ Cf. also Tzetzes, Chil. x11. 270, τῶν ᾿Αρχιμήδους μηχανών χρείαν ἔχω. 


CHAPTER. 1|. 


MANUSCRIPTS AND PRINCIPAL EDITIONS—ORDER OF 
COMPOSITION—-DIALECT—-LOST WORKS. 


THE sources of the text and versions are very fully described 
by Heiberg in the Prolegomena to Vol. 111. of his edition of Archi- 
medes, where the editor supplements and to some extent amends 
what he had previously written on the same subject in his dis- 
sertation entitled Quaestiones Archimedeae (1879). It will there- 
fore suffice here to state briefly the main points of the discussion. 

The MSS. of the best class all had a common origin in a MS. 
which, so far as is known, is no longer extant. It is described 
in one of the copies made from it (to be mentioned later and dating 
from some time between A.D. 1499 and 1531) as ‘most ancient’ 
(παλαιοτάτου), and all the evidence goes to show that it was written 
as early as the 9th or 10th century. At one time it was in the 
possession of George Valla, who taught at Venice between the 
years 1486 and 1499; and many important inferences with regard 
to its readings can be drawn from some translations of parts of 
Archimedes and Eutocius made by Valla himself and published 
in his book entitled de expetendis et fugiendis rebus (Venice, 1501). 
It appears to have been carefully copied from an original belonging 
to some one well versed in mathematics, and it contained figures 
drawn for the most part with great care and accuracy, but there 
was considerable confusion between the letters in the figures and 
those in the text. This MS., after the death of Valla in 1499, 
became the property of Albertus Pius Carpensis (Alberto Pio, 
prince of Carpi).. Part of his library passed through various hands 
and ultimately reached the Vatican; but the fate of the Valla 
MS. appears to have been different, for we hear of its being in 
the possession of Cardinal Rodolphus Pius (Rodolfo Pio), a nephew 
of Albertus, in 1544, after which it seems to have disappeared. 


XX1V INTRODUCTION. 


The three most important MSS. extant are: 


F (=Codex Florentinus bibliothecae Laurentianae Mediceae 
plutei xxvii. 4to.). 


B (=Codex Parisinus 2360, olim Mediceus). 
C (=Codex Parisinus 2361, Fonteblandensis). 


Of these it is certain that B was copied from the Valla MS, 
This is proved by a note on the copy itself, which states that the 
archetype formerly belonged to George Valla and afterwards to 
Albertus Pius. From this it may also be inferred that B was 
written before the death of Albertus in 1531; for, if at the date 
of B the Valla MS. had passed to Rodolphus Pius, the name of 
the latter would presumably have been mentioned. The note re- 
ferred to also gives a list of peculiar abbreviations used in the 
archetype, which list is of importance for the purpose of com- 
parison with F and other MSS. 

From a note on C it appears that that MS. was written by 
one Christophorus Auverus at Rome in 1544, at the expense of 
Georgius Armagniacus (Georges d’Armagnac), Bishop of Rodez, 
then on a mission from King Francis I. to Pope Paul III. Further, 
a certain Guilelmus Philander, in a letter to Francis I. published 
in an edition of Vitruvius (1552), mentions that he was allowed, 
by the kindness of Cardinal Rodolphus Pius, acting at the instance 
of Georgius Armagniacus, to see and make extracts from a volume 
of Archimedes which was destined to adorn the library founded 
by Francis at Fontainebleau. He adds that the volume had been 
the property of George Valla. We can therefore hardly doubt 
that C was the copy which Georgius Armagniacus had made in 
order to present it to the library at Fontainebleau. 

Now F, B and C all contain the same works of Archimedes 
and Eutocius, and in the same order, viz. (1) two Books de sphaera 
et cylindro, (2) de dimensione circuli, (3) de conoidibus, (4) de 
lineis spiralibus, (5) de planis aeque ponderantibus, (6) arenarius, 
(7) quadratura parabolae, and the commentaries of Eutocius on 
(1) (2) and (5). At the end of the quadratura parabolae both 
F and B give the following lines: 

εὐτυχοίης λέον γεώμετρα 

πολλοὺς εἰς λυκάβαντας ἴοις πολὺ φίλτατε μούσαις. 
F and C also contain menswrae from Heron and two fragments 
περὶ σταθμῶν and περὶ μέτρων, the order being the same in both 


MANUSCRIPTS. XXV 


and the contents only differing in the one respect that the last 
fragment περὶ μέτρων is slightly longer in F than in C. 

A short preface to C states that the first page of the archetype 
was so rubbed and worn with age that not even the name of 
Archimedes could be read upon it, while there was no copy at 
Rome by means of which the defect could be made good, and 
further that the last page of Heron’s de mensuris was similarly 
obliterated. Now in F the first page was apparently left blank 
at first and afterwards written in by a different hand with many 
gaps, while in B there are similar deficiencies and a note attached 
by the copyist is to the effect that the first page of the archetype 
was indistinct. In another place (p. 4 of Vol. πι., ed. Heiberg) 
all three MSS. have the same lacuna, and the scribe of B notes 
that one whole page or even two are missing. 

Now C could not have been copied from F because the last 
page of the fragment περὶ μέτρων is perfectly distinct in F; and, 
on the other hand, the archetype of F must have been illegible 
at the end because there is no word τέλος at the end of F, nor any 
other of the signs by which copyists usually marked the completion 
of their task. Again, Valla’s translations show that his MS. had 
certain readings corresponding to correct readings in B and C 
instead of incorrect readings given by F. Hence F cannot have 
been. Valla’s MS. itself. 

The positive evidence about F is as follows. Valla’s trans- 
lations, with the exception of the few readings just referred to, 
agree completely with the text of F. From a letter written at 
Venice in 1491 by Angelus Politianus (Angelo Poliziano) to Lau- 
rentius Mediceus (Lorenzo de’ Medici), it appears that the former 
had found a MS. at Venice containing works by Archimedes and 
Heron and proposed to have it copied. As G. Valla then lived 
at Venice, the MS. can hardly have been any other but his, and 
no doubt F was actually copied from it in 1491 or soon after. 
Confirmatory evidence for this origin of F is found in the fact 
that the form of most of the letters in it is older than the 15th 
century, and the abbreviations etc., while they all savour of an 
ancient archetype, agree marvellously with the description which 
the note to B above referred to gives of the abbreviations used 
in Valla’s MS. Further, it is remarkable that the corrupt passage 
corresponding to the illegible first page of the archetype just takes 
up one page of F, no more and no less. 


XXV1 INTRODUCTION. 


The natural inference from all the evidence is that F, B and 
C all had their origin in the Valla MS.; and of the three F is 
the most trustworthy. For (1) the extreme care with which the 
copyist of F kept to the original is illustrated by a number of 
mistakes in it which correspond to Valla’s readings but are cor- | 
rected in B and C, and (2) there is no doubt that the writer of 
B was somewhat of an expert and made many alterations on his 
own authority, not always with success. 

Passing to other MSS., we know that Pope Nicholas V. had 
a MS. of Archimedes which he caused to be translated into Latin. 
The translation was made by Jacobus Cremonensis (Jacopo Cas- 
siani*), and one copy of this was written out by Joannes Regio- 
montanus (Johann Miiller of Koénigsberg, near Hassfurt, in Fran- 
conia), about 1461, who not only noted in the margin a number 
of corrections of the Latin but added also in many places Greek 
readings from another MS. This copy by Regiomontanus is pre- 
served at Niirnberg and was the source of the Latin translation 
given in the editio princeps of Thomas Gechauff Venatorius (Basel, 
1544); it is called ΝΡ by Heiberg. (Another copy of the same 
translation is alluded to by Regiomontanus, and this is doubtless 
the Latin MS. 327 of 15th c. still extant at Venice.) From the 
fact that the translation of Jacobus Cremonensis has the same 
lacuna as that in F, B and C above referred to (Vol. 111, ed. 
Heiberg, p. 4), it seems clear that the translator had before him 
either the Valla MS. itself or (more likely) a copy of it, though 
the order of the books in the translation differs in one respect 
from that in our MSS., viz. that the arenarius comes after instead 
of before the guadratura parabolae. 

It is probable that the Greek MS. used by Regiomontanus was V 
(= Codex Venetus Marcianus cccy. of the 15th c.), which is still extant 
and contains the same books of Archimedes and Eutocius with the 
same fragment of Heron as F has, and in the same order. If the 
above conclusion that F dates from 1491 or thereabouts is correct, 
then, as V belonged to Cardinal Bessarione who died in 1472, it 
cannot have been copied from F, and the simplest way of accounting 
for its similarity to F is to suppose that it too was derived from 
Valla’s MS. 


* Tiraboschi, Storia della Letteratura Italiana, Vol. vi. Pt. 1 (p. 358 of the 
edition of 1807). Cantor (Vorlesungen iib. Gesch. d. Math., τα. p. 192) gives the 
full name and title as Jacopo da S. Cassiano Cremonese canonico regolare. 


MANUSCRIPTS. XXV11 


Regiomontanus mentions, in a note inserted later than the 
rest and in different ink, two other Greek MSS., one of which he 
calls “exemplar vetus apud magistrum Paulum.” Probably the 
monk Paulus (Albertini) of Venice is here meant, whose date was 
1430 to 1475; and it is possible that the ‘exemplar vetus” is 
the MS. of Valla. 

The two other inferior MSS., viz. A (=Codex Parisinus 2359, 
olim Mediceus) and D (=Cod. Parisinus 2362, Fonteblandensis), 
owe their origin to V. 

It is next necessary to consider the probabilities as to the MSS. 
used by Nicolas Tartaglia for his Latin translation of certain of 
the works of Archimedes. The portion of this translation published 
at Venice in 1543 contained the books de centris gravium vel de 
aequerepentibus I-II, tetragonismus | parabolae|, dimensio circuli 
and de insidentibus aquae I; the rest, consisting of Book II de 
insidentibus aquae, was published with Book I of the same treatise, 
after Tartaglia’s death in 1557, by Troianus Curtius (Venice, 1565), 
Now the last-named treatise is not extant in any Greek MS. and, 
as Tartaglia adds it, without any hint of a separate origin, to the 
rest of the books which he says he took from a mutilated and 
almost illegible Greek MS., it might easily be inferred that the 
Greek MS. contained that treatise also. But it is established, by 
a letter written by Tartaglia himself eight years later (1551) that 
he then had no Greek text of the Books de insidentibus aquae, and 
it would be strange if it had disappeared in so short a time without 
leaving any trace. Further, Commandinus in the preface to his 
edition of the same treatise (Bologna, 1565) shows that he had 
never heard of a Greek text of it. Hence it is most natural to 
suppose that it reached Tartaglia from some other source and in the 
Latin translation only*. 

The fact that Tartaglia speaks of the old MS. which he used 
as “fracti et qui vix legi poterant libri,” at practically the same 
time as the writer of the preface to C was giving a similar de- 
scription of Valla’s MS., makes it probable that the two were 


* The Greek fragment of Book 1., περὶ τῶν ὕδατι ἐφισταμένων ἢ περὶ τῶν 
ὀχουμένων, edited by A. Mai from two Vatican MSS. (Classici auct. 1. p. 426-30 ; 
Vol. τι. of Heiberg’s edition, pp. 356-8), seems to be of doubtful authenticity. 
Except for the first proposition, it contains enunciations only and no proofs. 
Heiberg is inclined to think that it represents an attempt at retranslation into 
Greek made by some mediaeval scholar, and he compares the similar attempt 
made by Rivault. 


XXV1i1 INTRODUCTION. 


identical ; and this probability is confirmed by a considerable agree- 
ment between the mistakes in Tartaglia and in Valla’s versions. 

But in the case of the quadratura parabolae and the dimensio 
circuli Tartaglia adopted bodily, without alluding in any way to 
the source of it, another Latin translation published by Lucas’ 
Gauricus “Tuphanensis ex regno Neapolitano” (Luca Gaurico of 
Gifuni) in 1503, and he copied it so faithfully as to reproduce most 
obvious errors and perverse punctuation, only filling up a few 
gaps and changing some figures and letters. This translation by 
Gauricus is seen, by means of a comparison with Valla’s readings 
and with the translation of Jacobus Cremonensis, to have been 
made from the same MS. as the latter, viz. that of Pope Nicolas V. 

Even where Tartaglia used the Valla MS. he does not seem 
to have taken very great pains to decipher it when it was 
not easily legible—it may be that he was unused to deciphering 
MSS.—-and in such cases he did not hesitate to draw from other 
sources. In one place (de planor. equilib. 1. 9) he actually 
gives as the Archimedean proof a paraphrase of Eutocius some- 
what retouched and abridged, and in many other instances he 
has inserted corrections and interpolations from another Greek 
MS. which he once names. This MS. appears to have been a copy 
made from F, with interpolations due to some one not unskilled 
in the subject-matter; and this interpolated copy of F was ap- 
parently also the source of the Nirnberg MS. now to be mentioned. 

N* (= Codex Norimbergensis) was written in the 16th century 
and brought from Rome to Niirnberg by Wilibald Pirckheymer. 
It contains the same works of Archimedes and Eutocius, and in 
the same order, as F, but was evidently not copied from F direct, 
while, on the other hand, it agrees so closely with Tartaglia’s 
version as to suggest a common origin. N* was used by Vena- 
torius in preparing the editio princeps, and Venatorius corrected 
many mistakes in it with his own hand by notes in the margin 
or on slips attached thereto; he also made many alterations in 
the body of it, erasing the original, and sometimes wrote on it 
directions to the printer, so that it was probably actually used 
to print from. The character of the MS. shows it to belong to 
the same class as the others; it agrees with them in the more 
important errors and in having a similar lacuna at the beginning. 
Some mistakes common to it and F alone show that its source was 
F, though at second hand, as above indicated. 


EDITIONS AND TRANSLATIONS. XX1X 


It remains to enumerate the principal editions of the Greek 
text and the published Latin versions which are based, wholly or 
partially, upon direct collation of the MSS. These are as follows, 
in addition to Gaurico’s and Tartaglia’s translations. 


1. The editio princeps published at Basel in 1544 by Thomas 
Gechauff Venatorius under the title Archimedis opera quae quidem 
exstant omnia nunc primum graece et latine in lucem edita. Advecta 
quoque sunt Eutocit Ascalonitae commentaria item graece et latine 
nunquam antea excusa. The Greek text and the Latin version in 
this edition were taken from different sources, that of the Greek 
text being N*, while the translation was Joannes Regiomontanus’ 
revised copy (N?) of the Latin version made by Jacobus Cremo- 
nensis from the MS. of Pope Nicolas V. The revision by 
Regiomontanus was effected by the aid of (1) another copy of 
the same translation still extant, (2) other Greek MSS., one of 
which was probably V, while another may have been Valla’s MS. 
itself. 


2. A translation by F. Commandinus (containing the following 
works, corcult dimensio, de lineis sprralibus, quadratura parabolae, 
de conoidibus et sphaerordibus, de arenae numero) appeared at 
Venice in 1558 under the title Archimedis opera nonnulla in 
latinum conversa et commentaris illustrata. For this translation 
several MSS. were used, among which was V, but none preferable 
to those which we now possess. 


3. D. Rivault’s edition, Archimedis opera quae exstant graece 
et latine novis demonstr. et comment. rllustr. (Paris, 1615), gives 
only the propositions in Greek, while the proofs are in Latin and 
somewhat retouched. Rivault followed the Basel editio princeps 
with the assistance of B. 


4, Torelli’s edition (Oxford, 1792) entitled ᾿Αρχιμήδους τὰ σω- 
ζόμενα peta τῶν Εὐτοκίου ᾿Ασκαλωνίτου ὑπομνημάτων, Archimedis 
quae supersunt omnia cum Eutoci Ascalonitae commentariis ex 
recensione J. Torelli Veronensis cum nova versione latina.  <Acced- 
unt lectiones variantes ex codd. Mediceo et Parisiensibus. Torelli 
followed the Basel editio princeps in the main, but also collated 
V. The book was brought out after Torelli’s death by Abram 
Robertson, who added the collation of five more MSS., F, A, B, C, D, 
with the Basel edition. The collation however was not well done, 
and the edition was not properly corrected when in the press. 


XXX INTRODUCTION. 


5. Last of all comes the definitive edition of Heiberg (Archi- 
medis opera omnia cum commentariis Eutoci. E codice Florentino 
recensuit, Latine uertit notisque ilustrauit J. L. Heiberg. Leipzig, 
1880—1). 

The relation of all the MSS. and the above editions and trans- 
lations is well shown by Heiberg in the following scheme (with 
the omission, however, of his own edition) : 


Codex Uallae saec. 1x—x 


Cod. Nicolai V F Tartalea Vv B ee 
6. 1453 e. 1491 a. 1543 saec. Xv c. 1500 a. 1544 
ΕΞ 5555." 55. 
Cod. Tartaleae 11 | Ed. Riualti 
a. 1615 
N® saec. xvi mee 2 ΕΘΘΕ = 
| Age) Commandinus 
Ed. Basil. 1544 saec. XVI 1558 
Torellius 1792 
a oS 
Gauricus Cremonensis 6. 1460 
fp Re ee 
Cod. Uenet. 827 ΝΡ, c. 1461 
5866. Xv 


The remaining editions which give portions of Archimedes in 
Greek, and the rest of the translations of the complete works or 
parts of them which appeared before Heiberg’s edition, were not 
based upon any fresh collation of the original sources, though some 
excellent corrections of the text were made by some of the editors, 
notably Wallis and Nizze. The following books may be mentioned. 


Joh, Chr. Sturm, Des unvergleichlichen Archimedis Kunstbiicher, 
tibersetzt und erldutert (Nirnberg, 1670). This translation em- 
braced all the works extant in Greek and followed three years 
after the same author’s separate translation of the Sand-reckoner. 
It appears from Sturm’s preface that he principally used the edition 
of Rivault. 


Is. Barrow, Opera Archimedis, Apollonii Pergaei conicorum libri, 
Theodosii sphaerica methodo novo illustrata et demonstrata (London, 


1675). 


Wallis, Archimedis arenarius et dimensio circuli, Eutocii in hanc 
commentary cum versione et notis (Oxford, 1678), also given 
in Wallis’ Opera, Vol. ur. pp. 509—546, 


Karl Friedr. Hauber, Archimeds zwei Biicher viber Kugel und 


Cylinder. Ebendesselben Kreismessung. Uebersetzt mit Anmerkungen 
u. 8. w. begleitet (Tubingen, 1798). 


TRANSLATIONS—ORDER OF WORKS. ἘΣΣῚ 


F. Peyrard, Quvres αἱ Archiméde, traduites littéralement, avec 
un commentaire, suivies dun mémoire du traducteur, sur un nouveau 
miroir ardent, et d'un autre mémoire de M. Delambre, sur Varith- 
métique des Grecs. (Second edition, Paris, 1808.) 

Ernst Nizze, Archimedes von Syrakus vorhandene Werke, aus dem 
Griechischen tibersetzt und mit erlauternden und kritischen Anmer- 
kungen begleitet (Stralsund, 1824). 

The MSS. give the several treatises in the following order. 

1. περὶ σφαίρας καὶ κυλίνδρου α΄ β΄, two Books On the Sphere 
and Cylinder. 
κύκλου μέτρησις ἢ, Measurement of a Circle. 
περὶ κωνοειδέων Kal σφαιροειδέων, On Conoids and Spheroids. 


περὶ ἑλίκων, On Spirals. 


or HR ὧν bo 


ἐπιπέδων ἰσορροπιῶν a Bt, two Books On the Hquwilibrium 
of Planes. 

6. ψαμμίτης, The Sand-reckoner. 

7. τετραγωνισμὸς παραβολῆς (a name substituted later for that 
given to the treatise by Archimedes himself, which must 
undoubtedly have been τετραγωνισμὸς τῆς τοῦ ὀρθογωνίου 


κώνου τομῆς 1), Quadrature of the Parabola. 
To these should be added 


8. περὶ ὀχουμένων ὃ, the Greek title of the treatise On floating 
bodies, only preserved in a Latin translation. 


* Pappus alludes (1. p. 312, ed. Hultsch) to the κύκλου μέτρησις in the words 
ἐν τῷ περὶ τῆς TOO κύκλου περιφερείας. 

+ Archimedes himself twice alludes to properties proved in Book 1. as 
demonstrated ἐν τοῖς μηχανικοῖς (Quadrature of the Parabola, Props. 6, 10). 
Pappus (στα. p. 1034) quotes τὰ ᾿Αρχιμήδους περὶ ἰσορροπιῶν. The beginning of 
Book 1. is also cited by Proclus in his Commentary on Eucl. 1., Ὁ. 181, where the 
reading should be τοῦ ἃ ἰσορροπιών, and not τῶν ἀνισορροπιῶν (Hultsch). 

+ The name ‘ parabola’ was first applied to the curve by Apollonius. Archi- 
medes always used the old term ‘section of a right-angled cone.’ Cf. Eutocius 
(Heiberg, vol. 111., p. 342) δέδεικται ἐν τῷ περὶ τῆς TOD ὀρθογωνίου κώνου τομῆς. 

§ This title corresponds to the references to the book in Strabo 1. p. 54 
(Ἀρχιμήδης ἐν τοῖς περὶ τῶν ὀχουμένων) and Pappus vir. p. 1024 (ὡς ᾿Αρχιμήδης 
ὀχουμένοι5). The fragment edited by Mai has a longer title, περὶ τῶν ὕδατι 
ἐφισταμένων ἢ περὶ τῶν ὀχουμένων, where the first part corresponds to Tartaglia’s 
version, de insidentibus aquae, and to that of Commandinus, de iis quae vehun- 
tur in aqua. But Archimedes intentionally used the more general word ὑγρόν 
(fluid) instead of ὕδωρ; and hence the shorter title περὶ ὀχουμένων, de iis quae 
in humido vehuntur (Torelli and Heiberg), seems the better. 


ΧΧΧΙΙ INTRODUCTION. 


The books were not, however, written in the above order; and 
Archimedes himself, partly through his prefatory letters and partly 
by the use in later works of properties proved in earlier treatises, 
gives indications sufficient to enable the chronological sequence 
to be stated approximately as follows : 


On the equilibrium of planes, I. 
2. Quadrature of the Parabola. 
3. On the equilibrium of planes, 11. 
4. On the Sphere and Cylinder, I, ΤΊ. 
5. On Spirals. 
6. On Conoids and Spheroids. 
7. On floating bodies, I, 11. 
8. Measurement of a circle. 
9. The Sand-reckoner. 


It should however be observed that, with regard to (7), no 
more is certain than that it was written after (6), and with regard 
to (8) no more than that it was later than (4) and before (9). 

In addition to the above we have a collection of Lemmas (Liber 
Assumptorum) which has reached us through the Arabic. The 
collection was first edited by 8. Foster, Atscellanea (London, 1659), 
and next by Borelli in a book published at Florence, 1661, in 
which the title is given as Liber assumptorum Archimedis interprete 
Thebit ben Kora et exponente doctore Almochtasso Abilhasan. The 
Lemmas cannot, however, have been written by Archimedes in 
their present form, because his name is quoted in them more than 
once. The probability is that they were propositions collected by 
some Greek writer* of a later date for the purpose of elucidating 
some ancient work, though it is quite likely that some of the 
propositions were of Archimedean origin, e.g. those concerning 
the geometrical figures called respectively ἀἄρβηλος ἡ (literally 


* Tt would seem that the compiler of the Liber Assumptorum must have 
drawn, to a considerable extent, from the same sources as Pappus. The 
number of propositions appearing substantially in the same form in both 
collections is, I think, even greater than has yet been noticed. Tannery (La 
Géométrie grecque, p. 162) mentions, as instances, Lemmas 1, 4, 5, 6; but it 
will be seen from the notes in this work that there are several other coin- 
cidences. 

+ Pappus gives (p. 208) what he calls an ‘ancient proposition’ (ἀρχαία 
mporacis) about the same figure, which he describes as χωρίον, ὃ δὴ καλοῦσιν 
ἄρβηλον. Cf. the note to Prop. 6 (p. 308). The meaning of the word is gathered 


WORKS ASCRIBED TO ARCHIMEDES. XXX111 


‘shoemaker’s knife’) and σάλινον (probably a ‘salt-cellar’*), and 
Prop. 8 which bears on the problem of trisecting an angle. 


from the Scholia to Nicander, Theriaca, 423: ἄρβηλοι λέγονται τὰ κυκλοτερῆ 
σιδήρια, οἷς οἱ σκυτοτόμοι τέμνουσι καὶ ξύουσι τὰ δέρματα. Cf. Hesychius, 
ἀνάρβηλα, τὰ μὴ ἐξεσμένα δέρματα" ἄρβηλοι γὰρ τὰ σμιλία. 

* The best authorities appear to hold that in any case the name σάλινον was 
not applied to the figure in question by Archimedes himself but by some later 
writer. Subject to this remark, I believe σάλινον to be simply a Graecised 
form of the Latin word salinwm. We know that a salt-cellar was an essential 
part of the domestic apparatus in Italy from the early days of the Roman 
Republic. ‘All who were raised above poverty had one of silver which 
descended from father to son (Hor., Carm. τι. 16, 13, Liv. xxvr. 36), and 
was accompanied by a silver patella which was used together with the salt- 
cellar in the domestic sacrifices (Pers. 11. 24, 25). These two articles of 
silver were alone compatible with the simplicity of Roman manners in the 
early times of the Republic (Plin., H. N. xxxiu. § 153, Val. Max. tv. 4, ὃ 3). 
...In shape the salinwm was probably in most cases a round shallow bowl” 
[Dict. of Greek and Roman Antiquities, article salinum]. Further we have 
in the early chapters of Mommsen’s History of Rome abundant evidence 
of similar transferences of Latin words to the Sicilian dialect of Greek. Thus 
(Book 1., ch. xiii.) it is shown that, in consequence of Latino-Sicilian com- 
merce, certain words denoting measures of weight, libra, triens, quadrans, 
seatans, uncia, found their way into the common speech of Sicily in the third 
century of the city under the forms λίτρα, τριᾶς, τετρᾶς, ἑξᾶς, οὐγκία. Similarly 
Latin law-terms (ch. xi.) were transferred; thus mutwum (a form of loan) 
became μοῖτον, carcer (a prison) kdpxapov. Lastly, the Latin word for lard, 
arvina, became in Sicilian Greek ἀρβίνη, and patina (a dish) πατάνη. The last 
word is as close a parallel for the supposed transfer of salinwm as could be 
wished. Moreover the explanation of σάλινον as salinum has two obvious 
advantages in that (1) it does not require any alteration in the word, and 


(2) the resemblance of the lower curve to an ordinary type of salt-cellar is 

evident. I should add, as confirmation of my hypothesis, that Dr A. S. Murray, 

of the British Museum, expresses the opinion that we cannot be far wrong in 

accepting as a salinuwm one of the small silver bowls in the Roman ministerium 
Hi. A. c 


XXX1V INTRODUCTION. 


Archimedes is further credited with the authorship of the 
Cattle-problem enunciated in the epigram edited by Lessing in 
1773. According to the heading prefixed to the epigram it was 
communicated by Archimedes to the mathematicians at Alexandria 
in a letter to Eratosthenes*. There is also in the Scholia to Plato’s 
Charmides 165 πὶ ἃ reference to the problem “called by Archimedes 
the Cattle-problem” (τὸ κληθὲν ὑπ᾽ ᾿Αρχιμήδους βοεικὸν πρόβλημα). 
The question whether Archimedes really propounded the problem, 
or whether his name was only prefixed to it in order to mark the 
extraordinary difficulty of it, has been much debated. A complete 
account of the arguments for and against is given in an article 
by Krumbiegel in the Zeitschrift fiir Mathematik und Physik 
(Hist. litt, Abtheilung) xxv. (1880), p. 121 sq., to which Amthor 
added (ibid. p. 153 sq.) a discussion of the problem itself. The . 
general result of Krumbiegel’s investigation is to show (1) that 


at the Museum which was found at Chaourse (Aisne) in France and is of a 
section sufficiently like the curve in the Salinon. 

The other explanations of σάλινον which have been suggested are as follows. 

(1) Cantor connects it with σάλος, ‘‘das Schwanken des hohen Meeres,” 
and would presumably translate it as wave-line. But the resemblance is 
not altogether satisfactory, and the termination -ἰνον would need explanation. 

(2) Heiberg says the word is ‘‘sine dubio ab Arabibus deprauatum,” and 
suggests that it should be σέλινον, parsley (‘‘ex similitudine frondis apii’’). 
But, whatever may be thought of the resemblance, the theory that the word is 
corrupted is certainly not supported by the analogy of ἄρβηλος which is correctly 
reproduced by the Arabs, as we know from the passage of Pappus referred to in 
the last note. 

(3) Dr Gow suggests that σάλινον may be a ‘sieve,’ comparing σάλαξ. But 
this guess is not supported by any evidence. 

* The heading is, Πρόβλημα ὅπερ ᾿Αρχιμήδης ἐν ἐπιγράμμασιν εὑρὼν τοῖς ἐν 
᾿Αλεξανδρείᾳ περὶ ταῦτα πραγματευομένοις ζητεῖν ἀπέστειλεν ἐν τῇ πρὸς ᾿Ἐρατοσθένην 
τὸν Κυρηναῖον ἐπιστολῇ. Heiberg translates this as ‘‘the problem which 
Archimedes discovered and sent in an epigram...in a letter to Eratosthenes.” 
He admits however that the order of words is against this, as is also the use of 
the plural ἐπιγράμμασιν. It is clear that to take the two expressions ἐν 
ἐπιγράμμασιν and ἐν ἐπιστολῇ as both following ἀπέστειλεν is very awkward. In 
fact there seems to be no alternative but to translate, as Krumbiegel does, in 
accordance with the order of the words, ‘‘a problem which Archimedes found 
among (some) epigrams and sent...in his letter to Eratosthenes”’ ; and this sense 
is certainly unsatisfactory. Hultsch remarks that, though the mistake πραγ- 
ματουμένοις for πραγματευομένοις and the composition of the heading as a whole 
betray the hand of a writer who lived some centuries after Archimedes, yet he 
must have had an earlier source of information, because he could hardly have 
invented the story of the letter to Eratosthenes. 


WORKS ASCRIBED TO ARCHIMEDES. XXXV 


the epigram can hardly have been written by Archimedes in its 
present form, but (2) that it is possible, nay probable, that the 
problem was in substance originated by Archimedes. Hultsch* has 
an ingenious suggestion as to the occasion of it. It is known that 
Apollonius in his ὠκυτόκιον had calculated a closer approximation to 
the value of z than that of Archimedes, and he must therefore have 
worked out more difficult multiplications than those contained in 
the Measurement of a circle. Also the other work of Apollonius 
on the multiplication of large numbers, which is partly preserved 
in Pappus, was inspired by the Sand-reckoner of Archimedes ; and, 
though we need not exactly regard the treatise of Apollonius as 
polemical, yet it did in fact constitute a criticism of the earlier 
book. Accordingly, that Archimedes should then reply with a 
problem which involved such a manipulation of immense numbers 
as would be difficult even for Apollonius is not altogether outside 
the bounds of possibility. And there is an unmistakable vein of 
satire in the opening words of the epigram “Compute the number 
of the oxen of the Sun, giving thy mind thereto, if thou hast a 
share of wisdom,” in the transition from the first part to the 
second where it is said that ability to solve the first part would 
entitle one to be regarded as “not unknowing nor unskilled in 
numbers, but still not yet to be numbered among the wise,” and 
again in the last lines. Hultsch concludes that in any case the 
problem is not much later than the time of Archimedes and dates 
from the beginning of the 2nd century B.c. at the latest. 

Of the extant books it is certain that in the 6th century A.D. 
only three were generally known, viz. On the Sphere and Cylinder, 
the Measurement of a circle, and On the equilibrium of planes. Thus 
Eutocius of Ascalon who wrote commentaries on these works only 
knew the Quadrature of the Parabola by name and had never seen 
it nor the book On Spirals. Where passages might have been 
elucidated by references to the former book, Eutocius gives ex- 
planations derived from Apollonius and other sources, and he 
speaks vaguely of the discovery of a straight line equal to the 
circumference of a given circle “by means of certain spirals,” 
whereas, if he had known the treatise On Spirals, he would have 
quoted Prop. 18. There is reason to suppose that only the three 
treatises on which Eutocius commented were contained in the 


* Pauly-Wissowa’s Real-Encyclopidie, τι. 1, pp. 534, 5. 


XXXV1 INTRODUCTION. 


ordinary editions of the time such as that of Isidorus of Miletus, 
the teacher of Eutocius, to which the latter several times alludes. 

In these circumstances the wonder is that so many more books 
have survived to the present day. As it is, they have lost to a 
considerable extent their original form. Archimedes wrote in the 
Dorie dialect*, but in the best known books (On the Sphere and 
Cylinder and the Measurement of a circle) practically all traces 
of that dialect have disappeared, while a partial loss of Doric forms 
has taken place in other books, of which however the Sand- 
reckoner has suffered least. Moreover in all the books, except the 
Sand-reckoner, alterations and additions were first of all made by 
an interpolator who was acquainted with the Doric dialect, and 
then, at a date subsequent to that of Eutocius, the book On the 
Sphere and Cylinder and the Measurement of a circle were completely 
recast. 

Of the lost works of Archimedes the following can be identified. 


1. Investigations relating to polyhedra are referred to by 
Pappus who, after alluding (v. p. 352) to the five regular polyhedra, 
gives a description of thirteen others discovered by Archimedes 
which are semi-regular, being contained by polygons equilateral 
and equiangular but not similar. 


2. A book of arithmetical content, entitled ἀρχαί Principles 
and dedicated to Zeuxippus. We learn from Archimedes himself 
that the book dealt with the naming of numbers (κατονόμαξις τῶν 
ἀριθμῶν)  ἃπα expounded a system of expressing numbers higher 


* Thus Eutocius in his commentary on Prop. 4 of Book 11. On the Sphere 
and Cylinder speaks of the fragment, which he found in an old book and which 
appeared to him to be the missing supplement to the proposition referred to, 
as ‘‘preserving in part Archimedes’ favourite Doric dialect’? (ἐν μέρει δὲ τὴν 
᾿Αρχιμήδει φίλην Δωρίδα γλῶσσαν ἀπέσωζον). From the use of the expression ἐν 
μέρει Heiberg concludes that the Doric forms had by the time of Eutocius 
begun to disappear in the books which have come down to us no less than in 
the fragment referred to. 

+ Observing that in all the references to this work in the Sand-reckoner 
Archimedes speaks of the naming of numbers or of numbers which are named or have 
their names (ἀριθμοὶ κατωνομασμένοι, τὰ ὀνόματα ἔχοντες, τὰν κατονομαξίαν ἔχοντες), 
Hultsch (Pauly-Wissowa’s Real-Encyclopidie, ττ. 1, p. 511) speaks οὗ κατονό- 
pasts τῶν ἀριθμῶν as the name of the work; and he explains the words τινὰς τῶν 
ἐν ἀρχαῖς <dpiOuGv> τῶν κατονομαξίαν ἐχόντων aS meaning ‘‘some of the 
numbers mentioned at the beginning which have a special name,” where ‘at 
the beginning” refers to the passage in which Archimedes first mentions τῶν 


LOST WORKS. XXXVil 


than those which could be expressed in the ordinary Greek no- 
tation. This system embraced all numbers up to the enormous 
figure which we should now represent by a 1 followed by 80,000 
billion ciphers ; and, in setting out the same system in the Sand- 
reckoner, Archimedes explains that he does so for the benefit of 
those who had not had the opportunity of seeing the earlier work 
addressed to Zeuxippus. 


3. περὶ ζυγῶν, On balances or levers, in which Pappus says (VIII. 
p. 1068) that Archimedes proved that “ greater circles overpower 
(κατακρατοῦσι) lesser circles when they revolve about the same 
centre.” It was doubtless in this book that Archimedes proved 
the theorem assumed by him in the Quadrature of the Parabola, 
Prop. 6, viz. that, if a body hangs at rest from a point, the centre 
of gravity of the body and the point of suspension are in the same 
vertical line. 


4, xevtpoBapixa, On centres of gravity. This work is mentioned 
by Simplicius on Aristot. de caelo 11. (Scholia in Arist. 508 a 30). 
Archimedes may be referring to it when he says (On the equilibrium 
of planes 1. 4) that it has before been proved that the centre of 
gravity of two bodies taken together lies on the line joining the 
centres of gravity of the separate bodies. In the treatise On 
floating bodies Archimedes assumes that the centre of gravity of a 
segment of a paraboloid of revolution is on the axis of the segment 
at a distance from the vertex equal to Zrds of its length. This 
may perhaps have been proved in the κεντροβαρικά, if it was 
not made the subject of a separate work. 

Doubtless both the περὶ ζυγῶν and the κεντροβαρικά preceded 
the extant treatise On the equilibrium of planes. 


5. κατοπτρικά, an optical work, from which Theon (on Ptolemy, 
Synt. 1. p. 29, ed. Halma) quotes a remark about refraction. 
Cf. Olympiodorus in Aristot. Meteor., τι. p. 94, ed. Ideler. 


ὑφ᾽ ἁμῶν κατωνομασμένων ἀριθμῶν καὶ ἐνδεδομένων ἐν τοῖς ποτὶ Ζεύξιππον γεγραμ- 
μένοις. But ἐν ἀρχαῖς seems a less natural expression for ‘‘at the beginning” 
than ἐν ἀρχῇ or κατ᾽ ἀρχάς would have been. Moreover, there being no 
participial expression except κατονομαξίαν ἐχόντων to be taken with ἐν ἀρχαῖς in 
this sense, the meaning would be unsatisfactory ; for the numbers are not 
named at the beginning, but only referred to, and therefore some word like 
εἰρημένων should have been used. For these reasons I think that Heiberg, 
Cantor and Susemihl are right in taking ἀρχαί to be the name of the treatise. 


XXXVIl1 INTRODUCTION. 


6. περὶ σφαιροποιΐας, On sphere-making, a mechanical work on 
the construction of a sphere representing the motions of the 
heavenly bodies as already mentioned (p. xxi). 


7. ἐφόδιον, a Method, noticed by Suidas, who says that Theo- 
dosius wrote a commentary on it, but gives no further information 
about it. 


8. According to Hipparchus Archimedes must have written 
on the Calendar or the length of the year (cf. p. xxi). 


Some Arabian writers attribute to Archimedes works (1) On 
a heptagon in a circle, (2) On circles touching one another, (3) On 
parallel lines, (4) On triangles, (5) On the properties of right- 
angled triangles, (6) a book of Data; but there is no confirmatory 
evidence of his having written such works. A book translated 
into Latin from the Arabic by Gongava (Louvain, 1548) and en- 
titled antiqui scriptoris de speculo comburente concavitatis parabolae 
cannot be the work of Archimedes, since it quotes Apollonius. 


CHAPTER III. 
THE RELATION OF ARCHIMEDES TO HIS PREDECESSORS. 


An extraordinarily large proportion of the subject matter of 
the writings of Archimedes represents entirely new discoveries of 
his own. Though his range of subjects was almost encyclopaedic, 
embracing geometry (plane and solid), arithmetic, mechanics, hydro- 
statics and astronomy, he was no compiler, no writer of text- 
books ; and in this respect he differs even from his great successor 
Apollonius, whose work, like that of Euclid before him, largely 
consisted of systematising and generalising the methods used, and 
the results obtained, in the isolated efforts of earlier geometers. 
There is in Archimedes no mere working-up of existing materials ; 
g, some definite addition to 
the sum of knowledge, and his complete originality cannot fail 


his objective is always some new thin 


to strike any one who reads his works intelligently, without any 
corroborative evidence such as is found in the introductory letters 
prefixed to most of them. These introductions, however, are emi- 
nently characteristic of the man and of his work ; their directness 
and simplicity, the complete absence of egoism and of any effort 
to magnify his own achievements by comparison with those of 
others or by emphasising their failures where he himself succeeded : 
all these things intensify the same impression. . Thus his manner 
is to state simply what particular discoveries made by his pre- 
decessors had suggested to him the possibility of extending them 
in new directions; e.g. he says that, in connexion with the efforts 
of earlier geometers to square the circle and other figures, it 
occurred to him that no one had endeavoured to square a parabola, 
and he accordingly attempted the problem and finally solved it. 
In like manner, he speaks, in the preface of his treatise On the 


xl] INTRODUCTION. 


Sphere and Cylinder, of his discoveries with reference to those 
solids as supplementing the theorems about the pyramid, the cone 
and the cylinder proved by Eudoxus. He does not hesitate to 
say that certain problems baffled him for a long time, and that 
the solution of some took him many years to effect; and in one 
place (in the preface to the book On Spirals) he positively insists, 
for the sake of pointing a moral, on specifying two propositions 
which he had enunciated and which proved on further investigation 
to be wrong. The same preface contains a generous eulogy of 
Conon, declaring that, but for his untimely death, Conon would 
have solved certain problems before him and would have enriched 
geometry by many other discoveries in the meantime. 

In some of his subjects Archimedes had no fore-runners, e.g. 
in hydrostatics, where he invented the whole science, and (so 
far as mathematical demonstration was concerned) in his me- 
chanical investigations. In these cases therefore he had, in laying 
the foundations of the subject, to adopt a form more closely re- 
sembling that of an elementary textbook, but in the later parts 
he at once applied himself to specialised investigations. 

Thus the historian of mathematics, in dealing with Archimedes’ 
obligations to his predecessors, has a comparatively easy task before 
him. But it is necessary, first, to give some description of the use 
which Archimedes made of the general methods which had found 
acceptance with the earlier geometers, and, secondly, to refer to 
some particular results which he mentions as having been previously 
discovered and as lying at the root of his own investigations, or 
which he tacitly assumes as known. 


81. Use of traditional geometrical methods. 


In my edition of the Conics of Apollonius*, I endeavoured, 
following the lead given in Zeuthen’s work, Die Lehre von den 
Kegelschnitten im Altertum, to give some account of what has been 
fitly called the geometrical algebra which played such an important 
part in the works of the Greek geometers. The two main methods 
included under the term were (1) the use of the theory of pro- 
portions, and (2) the method of application of areas, and it was 
shown that, while both methods are fully expounded in the Elements 
of Euclid, the second was much the older of the two, being 
attributed by the pupils of Eudemus (quoted by Proclus) to the 


* Apollonius of Perga, pp. ci sqq. 


RELATION OF ARCHIMEDES TO HIS PREDECESSORS. xli 


Pythagoreans. It was pointed out that the application of areas, 
as set forth in the second Book of Euclid and extended in the 
sixth, was made by Apollonius the means of expressing what he 
takes as the fundamental properties of the conic sections, namely 
the properties which we express by the Cartesian equations 
y? = px, 
y? = pa + Ea 

referred to any diameter and the tangent at its extremity as axes ; 
and the latter equation was compared with the results obtained in the 
27th, 28th and 29th Props. of Euclid’s Book v1, which are equivalent 
to the solution, by geometrical means, of the quadratic equations 


b 
an+ —2’?=D. 
(5 


It was also shown that Archimedes does not, as a rule, connect his 
description of the central conics with the method of application of 
areas, as Apollonius does, but that Archimedes generally expresses 
the fundamental property in the form of a proportion 


2 12 


LEME: 


CDi.) eae 


and, in the case of the ellipse, 
y? b? 
=u 


2.0, (ἢ 


where a, x, are the abscissae measured from the ends of the diameter 
of reference. . 

It results from this that the application of areas is of much less 
frequent occurrence in Archimedes than in Apollonius. It is 
however used by the former in all but the most general form. The 
simplest form of ‘applying a rectangle” to a given straight line 
which shall be equal to a given area occurs e.g. in the proposition On 
the equilibrium of Planes τι. 1; and the same mode of expression 
is used (as in Apollonius) for the property y’= pz in the parabola, 
px being described in Archimedes’ phrase as the rectangle “applied 
to” (παραπίπτον παρά) a line equal to p and “having at its width” 
(πλάτος ἔχον) the abscissa (x). Then in Props. 2, 25, 26, 29 of the 
book On Conoids and Spheroids we have the complete expression 
which is the equivalent of solving the equation 

ax+ οἷ = 6’, 


“let a rectangle be applied (to a certain straight line) exceeding by 


xii INTRODUCTION. 


a square figure (παραπεπτωκέτω χωρίον ὑπερβάλλον εἴδει τετραγώνῳ) 
and equal to (a certain rectangle).” Thus a rectangle of this sort 
has to be made (in Prop. 25) equal to what we have above called 
x.«, in the case of the hyperbola, which is the same thing as 
a(a+a) or ax+a*, where ὦ is the length of the transverse axis. 
But, curiously enough, we do not find in Archimedes the application 
of a rectangle “falling short by a square figure,” which we should 
obtain in the case of the ellipse if we substituted «(a—~) for x. a. 
In the case of the ellipse the area x.a, is represented (On Conoids 
and Spheroids, Prop. 29) as a gnomon which is the difference 
between the rectangle h.h, (where h, h, are the abscissae of the 
ordinate bounding a segment of an ellipse) and a rectangle applied 
to h,—h and exceeding by a square figure whose side is ἢ -- α ; and 
the rectangle h.h, is simply constructed from the sides ἡ, h,. Thus 
Archimedes avoids* the application of a rectangle falling short by a 
square, using for x. a, the rather complicated form 


h.h,—{(hy—h) (ἃ -- αν) + (h—-«)’}. 
It is easy to see that this last expression is equal to ἃ). a, for it 
reduces to 
h.h,— th, (h—«)—a(h—-2)} 
=a(h, +h)-2’, 
=ax—«x’, since h,+h=a, 
nee 

It will readily be understood that the transformation of rectangles 
and squares in accordance with the methods of Euclid, Book 11, is 
just as important to Archimedes as to other geometers, and there is 
no need to enlarge on that form of geometrical algebra. 

The theory of proportions, as expounded in the fifth and sixth 
Books of Euclid, including the transformation of ratios (denoted by 
the terms componendo, dividendo, etc.) and the composition or 
multiplication of ratios, made it possible for the ancient geometers 
to deal with magnitudes in general and to work out relations 
between them with an effectiveness not much inferior to that of 
modern algebra. Thus the addition and subtraction of ratios could 
be effected by procedure equivalent to what we should in algebra 


* The object of Archimedes was no doubt to make the Lemma in Prop. 2 
(dealing with the summation of a series of terms of the form a.rx + (rzx)?, where r 
successively takes the values 1, 2, 3,...) serve for the hyperboloid of revolution 
and the spheroid as well. 


RELATION OF ARCHIMEDES TO HIS PREDECESSORS. ΧΙΠῚ 


call bringing to a common denominator. Next, the composition or 
multiplication of ratios could be indefinitely extended, and hence 
the algebraical operations of multiplication and division found easy 
and convenient expression in the geometrical algebra. As a par- 
ticular case, suppose that there is a series of magnitudes in continued 
proportion (i.e. in geometrical progression) as dy, 4), @2, ... G,, 80 that 


GH % ὄἄ,- 


We have then, by multiplication, 


a ONE a ἕ a 

1 ; 
a *) 5 ὉΓ — = ch τεῦ 3 
AX X X hy 


It is easy to understand how powerful such a method as that of 
proportions would become in the hands of an Archimedes, and a few 
instances are here appended in order to illustrate the mastery with 
which he uses it. 


1. A good example of a reduction in the order of a ratio after 
the manner just shown is furnished by On the equilibrium of Planes 
u1.10. Here Archimedes has a ratio which we will call a?/b*, where 
a?/6°=c/d; and he reduces the ratio between cubes to a ratio 
between straight lines by taking two lines a, y such that 


Ce τὰ 
"ἢ. (ἢ 
2 2 
It follows from this that (<) ee τς 
x Gao 
or te GS 
ὃ x’ 
a? ON oe a ὁ 
and hence m= (5) Seta: 


2. In the last example we have an instance of the use of 
auxiliary fixed lines for the purpose of simplifying ratios and 
thereby, as it were, economising power in order to grapple the more 
successfully with a complicated problem. With the aid of such 
auxiliary lines or (what is the same thing) auxiliary fixed points in 
a figure, combined with the use of proportions, Archimedes is able to 
effect some remarkable eliminations. 

Thus in the proposition On the Sphere and Cylinder τι. 4 he obtains 
three relations connecting three as yet undetermined points, and 


xliv INTRODUCTION. 


proceeds at once to eliminate two of the points, so that the problem 
is then reduced to finding the remaining point by means of one 
equation. Expressed in an algebraical form, the three original 
relations amount to the three equations 


θα--α y 


and the result, after the elimination of y and 2, is stated by 
Archimedes in a form equivalent to 


M+n A+2 4a? 
n  @ (2a—2)° 


Again the proposition On the equilibrium of Planes τι. 9 proves 
by the same method of proportions that, if a, ὦ, c, d, x, y, are straight 
lines satisfying the conditions 


“ἘΠ: Φ (a>b>e>d) 


Ben ἢ 

@ ῸΝ x 

a-d 3(a-c)’ 
2a+4b+6c+3d = y 
δα - 106 Ὁ 106- δα a—c’ 


then x+y = 2a. 


and 


The proposition is merely brought in as a subsidiary lemma to the 
proposition following, and is not of any intrinsic importance ; but a 
glance at the proof (which again introduces an auxiliary line) will 
show that it is a really extraordinary instance of the manipulation 
of proportions. 


3. Yet another instance is worth giving here. It amounts to 
the proof that, if 


ey 
—+ 5=1 
Gen dar oe 
2a+a 2a—a2 
then .y? (a—a)+ . 3 (a+2) — 4ab?. 
a+n ὅτι ) α--α γι ) 


A, A’ are the points of contact of two parallel tangent planes to a 
spheroid ; the plane of the paper is the plane through Ad’ and the 


RELATION OF ARCHIMEDES TO HIS PREDECESSORS. xlv 


axis of the spheroid, and PP’ is the intersection of this plane with 
another plane at right angles to it (and therefore parallel to the 
tangent planes), which latter plane divides the spheroid into two 
segments whose axes are AV, A’. Another plane is drawn through 


the centre and parallel to the tangent plane, cutting the spheroid 
into two halves. Lastly cones are drawn whose bases are the 
sections of the spheroid by the parallel planes as shown in the 
figure. 

Archimedes’ proposition takes the following form [On Conoids 
and Spheroids, Props. 31, 32]. 

APP’ being the smaller segment of the two whose common base 
is the section through PP’, and x, y being the coordinates of P, 
he has proved in preceding propositions that 


(volume of) segment APP’ 2a+¢a 


(volume of) cone APP’ = Aes a ae (a), 
half spheroid ABB’ 
= 26) 
and eae ABB’ Dy reece e rece eeccorece nae (3); 


and he seeks to prove that 
segment A’PP! _ 2a -« 
cone APE aa.” 
The method is as follows. 
cone ABB’ a δ a a 


We have : SS —— : : 
cone APP’ α-α y a-x @-# 
ΖΦ a 
If we suppose See ΓΕ IS αὐ 0 Oates eater 
PP ee (7), 
za 


the ratio of the cones becomes 


ea 


xlvi INTRODUCTION. 


Next, by hypothesis (a), 
cone APP’ _ a+a 
segemt. APP’ 2a+a° 


Therefore, ex aequali, 
cone ABB’ 2a 


segmt. APP’ (α-- α) (2a+2)’ 
It follows from (8) that 


spheroid 4ξα 


segmt. APP’ (a—2) (2a+2)’ 


Mee segmt. A'PP’ 4za—(a—2x) (2a+ 2) 
segmt. APP’ Ἕώ(α-- &) (2a +2) 


2 (2a—a)+(2a+a)(z-a—«x) 

(a—«) (2a + 2) ; 
Now we have to obtain the ratio of the segment A’PP’ to the cone 
A’PP', and the comparison between the segment APP’ and the cone 
A’PP’ is made by combining two ratios ex aequali. Thus 


segmt. APP’ Ἢ 9ᾳ +a (a) 
Τα ΣΡ a) aki a 


1 cone APP a—@ 
and ——— : 
cone A’PP’ a+2 


Thus combining the last three proportions, ex aequali, we have 


segmt. A’PP'  2z(2a—ax)+(2a+a) (6 --α -- αἡ 
cone A’PP’ — a? + 2ax + x 


_ 2#(2a— w)+(2a+2)(2-a-a a) 


z(a—x)+(2a+x) α : 


since a =2z(a—2), by (y). 


[The object of the transformation of the numerator and denominator 
of the last fraction, by which z(2a—«) and z(a—«) are made the 


a is the fraction which 


2a 
first terms, is now obvious, because 


Archimedes wishes to arrive at, and, in order to prove that the 
required ratio is equal to this, it is only necessary to show that 
2a-a@_ 6- z—(a—2) ἢ 
a—x a 


RELATION OF ARCHIMEDES TO HIS PREDECESSORS. xlvii 


Now SE ways zz 
a—x α--α 
z 
Ses: by (y), 
τᾷ 
τ ἢ 
ἘΞ Ὁ - 


ΠῚ 


ΠΕ segmt. A’PP’ 2ᾳ.--ὸὑψ 
ὴ ΘΟ ΠΟ a= ae 


4, One use by Euclid of the method of proportions deserves 
mention because Archimedes does not use it in similar circumstances. 
Archimedes (Quadrature of the Parabola, Prop. 23) sums a particular 
geometric series 

a+a(4+)+a(4)+...4+a(4)" 
in a manner somewhat similar to that of our text-books, whereas 
Euclid (1x. 35) sums any geometric series of any number of terms by 
means of proportions thus. 


Suppose aj, dy, ...@n, Gn, to be (n+1) terms of a geometric 
series in which a@,,, is the greatest term. Then 


nit my Uma 
On An_1 An_» Ay 
Therefore TEL Oe es an 
On, Gn _1 ay 


Adding all the antecedents and all the consequents, we have 


Anyi — GY Ay — Ay 


Ay + Ay + Ag + ...°+ Ay ay 


? 


which gives the sum of 7 terms of the series. 


§2. Harlier discoveries affecting quadrature and cuba- 
ture. 


Archimedes quotes the theorem that circles are to one another as 
the squares on their diameters as having being proved by earlier 
geometers, and he also says that it was proved by means of a certain 
lemma which he states as follows: “Of unequal lines, unequal 
surfaces, or unequal solids, the greater exceeds the less by such a 
magnitude as is capable, if added [continually] to itself, of exceeding 


xlvi INTRODUCTION. 


any given magnitude of those which are comparable with one another 
(τῶν πρὸς ἄλληλα λεγομένων). We know that Hippocrates of Chios 
proved the theorem that circles are to one another as the squares on 
their diameters, but no clear conclusion can be established as to the 
method which he used. On the other hand, Eudoxus (who is 
mentioned in the preface to The Sphere and Cylinder as having 
proved two theorems in solid geometry to be mentioned presently) 
is generally credited with the invention of the method of exhaustion 
by which Euclid proves the proposition in question in x11. 2. The 
lemma stated by Archimedes to have been used in the original proof 
is not however found in that form in Euclid and is not used in the 
proof of xit. 2, where the lemma used is that proved by him in 
x. 1, viz. that “Given two unequal magnitudes, if from the greater 
[a part] be subtracted greater than the half, if from the remainder 
[ἃ part] greater than the half be subtracted, and so on continually, 
there will be left some magnitude which will be less than the lesser 
given magnitude.” This last lemma is frequently assumed by 
Archimedes, and the application of it to equilateral polygons in- 
scribed in a circle or sector in the manner of x11. 2 is referred to as 
having been handed down in the Hlements*, by which it is clear 
that only Euclid’s Hlements can be meant. The apparent difficulty 
caused by the mention of éwo lemmas in connexion with the theorem 
in question can, however, I think, be explained by reference to 
the proof of x. 1 in Euclid. He there takes the lesser magnitude 
and says that it is possible, by multiplying it, to make it some time 
exceed the greater, and this statement he clearly bases on the 4th 
definition of Book vy. to the effect that ‘‘ magnitudes are said to bear 
a ratio to one another, which can, if multiplied, exceed one another.” 
Since then the smaller magnitude in x. 1 may be regarded as the 
difference between some two unequal magnitudes, it is clear that the 
lemma first quoted by Archimedes is in substance used to prove the 
lemma in x. 1 which appears to play so much larger a part in the in- 
vestigations in quadrature and cubature which have come down to us. 

The two theorems which Archimedes attributes to Eudoxus 
by namet are 


(1) that any pyramid is one third part of the prism which has 


the same base as the pyramid and equal height, and 


* On the Sphere and Cylinder, τ. 6. 
+ ibid. Preface. 


RELATION OF ARCHIMEDES TO HIS PREDECESSORS.  xlix 


(2) that any cone is one third part of the cylinder which has 
the same base as the cone and equal height. 


The other theorems in solid geometry which Archimedes quotes 
as having been proved by earlier geometers are*: 

(3) Cones of equal height are in the ratio of their bases, and 
conversely. 

(4) If a cylinder be divided by a plane parallel to the base, 
cylinder is to cylinder as axis to axis. 

(5) Cones which have the same bases as cylinders and equal 
height with them are to one another as the cylinders. 


(6) The bases of equal cones are reciprocally proportional to 
their heights, and conversely. 


(7) Cones the diameters of whose bases have the same ratio as 
their axes are in the triplicate ratio of the diameters of their bases. 


In the preface to the Quadrature of the Parabola he says 
that earlier geometers had also proved that 


(8) Spheres have to one another the triplicate ratio of their 
diameters ; and he adds that this proposition and the first of those 
which he attributes to Eudoxus, numbered (1) above, were proved 
by means of the same lemma, viz. that the difference between 
any two unequal magnitudes can be so multiplied as to exceed 
any given magnitude, while (if the text of Heiberg is right) the 
second of the propositions of Eudoxus, numbered (2), was proved 
by means of “a lemma similar to that aforesaid.” As a matter 
of fact, all the propositions (1) to (8) are given in Euclid’s twelfth 
Book, except (5), which, however, is an easy deduction from (2) ; 
and (1), (2), (3), and (7) all depend upon the same lemma [x. 1] 
as that used in Eucl. xi. 2. 

The proofs of the above seven propositions, excluding (5), as 
given by Euclid are too long to quote here, but the following sketch 
will show the line taken in the proofs and the order of the propo- 
sitions. Suppose ABCD to be a pyramid with a triangular base, 
and suppose it to be cut by two planes, one bisecting AB, AC, 
AD in F, G, FE respectively, and the other bisecting BC, BD, BA 
in H, K, F respectively. These planes are then each parallel to 
one face, and they cut off two pyramids each similar to the original 


* Lemmas placed between Props. 16 and 17 of Book 1. On the Sphere and 
Cylinder. 


H. A. d 


] INTRODUCTION. 


pyramid and equal to one another, while the remainder of the 
pyramid is proved to form two equal prisms which, taken together, 


A 


ef ee 
a πὰ 


are greater than one half of the original pyramid [x1. 3]. It is 
next proved [x11. 4] that, if there are two pyramids with triangular 
bases and equal height, and if they are each divided in the 
manner shown into two equal pyramids each similar to the whole 
and two prisms, the sum of the prisms in one pyramid is to the 
sum of the prisms in the other in the ratio of the bases of the 
whole pyramids respectively. Thus, if we divide in the same 
manner the two pyramids which remain in each, then all 
the pyramids which remain, and so on continually, it follows 
on the one hand, by x. 1, that we shall ultimately have 
pyramids remaining which are together less than any assigned 
solid, while on the other hand the sums of all the prisms 
resulting from the successive subdivisions are in the ratio of 
the bases of the original pyramids. Accordingly Euclid is able 
to use the regular method of exhaustion exemplified in xu. 2, 
and to establish the proposition [x11. 5] that pyramids with the 
same height and with triangular bases are to one another as their 
bases. The proposition is then extended [x1 6] to pyramids with the 
same height and with polygonal bases. Next [xt 7] a prism with 
a triangular base is divided into three pyramids which are shown 
to be equal by means of x11. 5; and it follows, as a corollary, that 
any pyramid is one third part of the prism which has the same 
base and equal height. Again, two similar and similarly situated 
pyramids are taken and the solid parallelepipeds are completed, 
which are then seen to be six times as large as the pyramids 
respectively ; and, since (by x1. 33) similar parallelepipeds are in 
the triplicate ratio of corresponding sides, it follows that the same 


D 


RELATION OF ARCHIMEDES TO HIS PREDECESSORS. li 


is true of the pyramids [x1. 8]. A corollary gives the obvious 
extension to the case of similar pyramids with polygonal bases, 
The proposition [xu. 9] that, in equal pyramids with triangular 
bases, the bases are reciprocally proportional to the heights is 
proved by the same method of completing the parallelepipeds and 
using x1. 34; and similarly for the converse. It is next proved 
[x11. 10] that, if in the circle which is the base of a cylinder a 
square be described, and then polygons be successively described 
by bisecting the arcs remaining in each case, and so doubling the 
number of sides, and if prisms of the same height as the cylinder 
be erected on the square and the polygons as bases respectively, 
the prism with the square base will be greater than half the 
cylinder, the next prism will add to it more than half of the 
remainder, and so on. And each prism is triple of the pyramid with 
the same base and altitude. Thus the same method of exhaustion 
as that in ΧΙ. 2 proves that any cone is one third part of the 
cylinder with the same base and equal height. Exactly the same 
method is used to prove [x1. 11] that cones and cylinders which 
have the same height are to one another as their bases, and 
[xi1. 12] that similar cones and cylinders are to one another in 
the triplicate ratio of the diameters of their bases (the latter 
proposition depending of course on the similar proposition xu. 8 
for pyramids). The next three propositions are proved without 
fresh recourse to x. 1. Thus the criterion of equimultiples laid 
down in Def. 5 of Book v. is used to prove [x11. 13] that, if a 
cylinder be cut by a plane parallel to its bases, the resulting 
cylinders are to one another as their axes. It is an easy deduction 
[x11. 14] that cones and cylinders which have equal bases are 
proportional to their heights, and [xu. 15] that in equal cones 
and cylinders the bases are reciprocally proportional to the heights, 
and, conversely, that cones or cylinders having this property are 
equal. Lastly, to prove that spheres are to one another in the 
triplicate ratio of their diameters [xu. 18], a new procedure is 
adopted, involving two preliminary propositions. In the first of 
these [xu. 16] it is proved, by an application of the usual lemma 
x. 1, that, if two concentric circles are given (however nearly 
equal), an equilateral polygon can be inscribed in the outer circle 
whose sides do not touch the inner ; the second proposition [x11. 17] 
uses the result of the first to prove that, given two concentric 
spheres, it is possible to inscribe a certain polyhedron in the outer 


ad 2 


ΠῚ INTRODUCTION. 


so that it does not anywhere touch the inner, and a corollary adds 
the proof that, if a similar polyhedron be inscribed in a second 
sphere, the volumes of the polyhedra are to one another in the 
triplicate ratio of the diameters of the respective spheres, This 
last property is then applied [x11. 18] to prove that spheres are 
in the triplicate ratio of their diameters. 


§ 3. Conic Sections. 


In my edition of the Conics of Apollonius there is a complete 
account of all the propositions in conics which are used by Archi- 
medes, classified under three headings, (1) those propositions 
which he expressly attributes to earlier writers, (2) those which 
are assumed without any such reference, (3) those which appear to 
represent new developments of the theory of conics due to Archi- 
medes himself. As all these properties will appear in this 
volume in their proper places, it will suffice here to state only 
such propositions as come under the first heading and a few under 
the second which may safely be supposed to have been previously 
known. 

Archimedes says that the following propositions “are proved 
in the elements of conics,” i.e. in the earlier treatises of Euclid 
and Aristaeus. 


1, In the parabola 


(a) if PV be the diameter of a segment and @Vq the 
chord parallel to the tangent at P, then YV=Vq; 


(Ὁ) if the tangent at ὦ meet VP produced in 7, then 
ey he ; 

(c) if two chords QVq, Q'V’q’ each parallel to the tangent 
at P meet the diameter PV in V, V’ respectively, 


PV: ΡΞ ΘΙ: 


2. If straight lines drawn from the same point touch any 
conic section whatever, and if two chords parallel to the respective 
tangents intersect one another, then the rectangles under the 
segments of the chords are to one another as the squares on the 
parallel tangents respectively. 


3. The following proposition is quoted as proved “in the conics.” 
If in a parabola p, be the parameter of the principal ordinates, 


RELATION OF ARCHIMEDES TO HIS PREDECESSORS. hii 


QQ’ any chord not perpendicular to the axis which is bisected in 1 
by the diameter PV, p the parameter of the ordinates to PV, and 
if QD be drawn perpendicular to PV, then 


OV? 3 OD =p: pa. 
[On Conoids and Spheroids, Prop. 3, which see. | 


The properties of a parabola, PV?=p,.AWN, and QV*=p.PYV, 
were already well known before the time of Archimedes. In fact 
the former property was used by Menaechmus, the discoverer of 
conic sections, in his duplication of the cube. 

It may be taken as certain that the following properties of the 
ellipse and hyperbola were proved in the Conices of Euclid. 


1. For the ellipse 
PN*?: AN. A'N=P'N": AN’. A’N'=CB? : CA’ 
and OV PVP Va OV ΡΣ ΒΕ OD) ΟἿ", 
(Either proposition could in fact be derived from the proposition 


about the rectangles under the segments of intersecting chords 
above referred to.) 


2. For the hyperbola 
AN PAW. 2 Vie PANN 1) 
and OW? PY OP! Viz Vite EV SOV 
though in this case the absence of the conception of. the double 
hyperbola as one curve (first found in Apollonius) prevented Euclid, 


and Archimedes also, from equating the respective ratios to those 
of the squares on the parallel semidiameters. 


3. In a hyperbola, if P be any point on the curve and PA, 
PL be each drawn parallel to one asymptote and meeting the 
other, 

PK. PL=(eonst.) 
This property, in the particular case of the rectangular hyperbola, 
was known to Menaechmus. 

It is probable also that the property of the subnormal of the 
parabola (VG'=4p,) was known to Archimedes’ predecessors. It 
is tacitly assumed, On floating bodies, τι. 4, etc. 

From the assumption that, in the hyperbola, 47’< AN (where 
N is the foot of the ordinate from P, and 7’ the point in which the 


liv INTRODUCTION. 


tangent at P meets the transverse axis) we may perhaps infer 
that the harmonic property 


GLY ES bat LT cee wal aria all 
or at least the particular case of it, 
DA NA 


was known before Archimedes’ time. 

Lastly, with reference to the genesis of conic sections from 
cones and cylinders, Euclid had already stated in his Phaenomena 
that, “if a cone or cylinder be cut by a plane not parallel to the 
base, the resulting section is a section of an acute-angled cone 
{an ellipse] which is similar to a @vpeds.” Though it is not probable 
that Euclid had in mind any other than a right cone, the statement 
should be compared with On Conoids and Spheroids, Props. 7, 8, 9. 


84. Surfaces of the second degree. 


Prop. 11 of the treatise On Conoids and Spheroids states without 
proof the nature of certain plane sections of the conicoids of revo- 
lution. Besides the obvious facts (1) that sections perpendicular 
to the axis of revolution are circles, and (2) that sections through 
the axis are the same as the generating conic, Archimedes asserts 
the following. 


1. In a paraboloid of revolution any plane section parallel to 
the axis is a parabola equal to the generating parabola. 


2. In a hyperboloid of revolution any plane section parallel 


to the axis is a hyperbola similar to the generating hyperbola. 


3. Ina hyperboloid of revolution a plane section through the 
vertex of the enveloping cone is a hyperbola which is not similar 
to the generating hyperbola. 


4. In any spheroid a plane section parallel to the axis is an 
ellipse similar to the generating ellipse. 


Archimedes adds that “the proofs of all these propositions 
are manifest (φανεραί). The proofs may in fact be supplied as 
follows. 

1. Section of a paraboloid of revolution by a plane parallel 
to the axis. 


RELATION OF ARCHIMEDES TO HIS PREDECESSORS. lv 


Suppose that the plane of the paper represents the plane section 
through the axis AV which intersects the given plane section at right 
angles, and let A’O be the line of intersection. 
Let POP’ be any double ordinate to AW in the 
section through the axis, meeting A’O and AV 
at right angles in O, WV respectively. Draw A’ 
perpendicular to AX. 

Suppose a perpendicular drawn from O to 
A'O in the plane of the given section parallel to 
the axis, and let y be the length intercepted by 
the surface on this perpendicular. 

Then, since the extremity of y is on the 
circular section whose diameter is PP’, 


y= PO.OP", 


If A’O =a, and if p is the principal parameter of the generating 
parabola, we have then 


y? = PN? ON? 
= PN? A'M? 
=p(AN-Al) 
= ρα, 


so that the section is a parabola equal to the generating parabola. 


2. Section of a hyperboloid of revolution by a plane parallel to 
the axis, 
Take, as before, the plane section through the axis which intersects 


the given plane section at right angles in A’O. Let the hyperbola 


lvi INTRODUCTION. 


PAP’ in the plane of the paper represent the plane section through 
the axis, and let C be the centre (or the vertex of the enveloping 
cone). Draw CC’ perpendicular to CA, and produce OA’ to meet it 
in C’. Let the rest of the construction be as before. 
Suppose that 
CA τ Ae" CO: 
and let y have the same meaning as before. 
Then P=PO OP ΞΟ =A: 
And, by the property of the original hyperbola, 
PN* : CN?-CA?=A'M? : CM? — CA? (which is constant). 
Thus «45.275: CM?—CA?= PWN? :CN?-CA? 
=PN?—-A'M’ : CN’-Cl? 
= Ὁ a οἷ - Ἴ 


whence it appears that the section is a hyperbola similar to the 
original one. 


3. Section of a hyperboloid of revolution by a plane passing 
through the centre (or the vertex of the enveloping cone). 


I think there can be no doubt that Archimedes would have proved 
his proposition about this section by means of the same general 
property of conics which he uses to prove Props. 3 and 12—14 of 
the same treatise, and which he enunciates at the beginning of 
Prop. 3 as a known theorem proved in the ‘elements of conics,” viz. 
that the rectangles under the segments of intersecting chords are as 
the squares of the parallel tangents. 

Let the plane of the paper represent the plane section through 
the axis which intersects the given plane passing through the 
centre at right angles. Let (/A’O be the line of intersection, C 
being the centre, and A’ being the point where ('A’O meets the 
surface. Suppose CAIN to be the axis of the hyperboloid, and 
POp, P'O'p' two double ordinates to it in the plane section through 
the axis, meeting C'A’O in O, O’ respectively ; similarly let 4΄ be 
the ordinate from A’. Draw the tangents at A and A’ to the 
section through the axis meeting in 7’, and let QOq, Q’O'q’ be the 
two double ordinates in the same section which are parallel to the 
tangent at A’ and pass through Ὁ, O’ respectively. 

Suppose, as before, that y, y’ are the lengths cut off by the 


RELATION OF ARCHIMEDES TO HIS PREDECESSORS. lvil 


surface from the perpendiculars at O and 0’ to OC in the plane of 
the given section through C'A’O, and that 


CO—a C0 =e. CA — nc A= 


Then, by the property of the intersecting chords, we have, since 
QO = 09, 
PO! Op 3Q0° — Ae PA 


Ξσ ca OGD; 2) OL OE 
Also PaO ρ ΞΡ Ὁ" 
and, by the property of the hyperbola, 
00? =r — a A= OF 5 — a: 
It follows, ex aequali, that 
Cia ert) an) ROY Pc nts el a ee ο (a), 
and therefore that the section is a hyperbola. 

To prove that this hyperbola is not similar to the generating 
hyperbola, we draw CC’ perpendicular to C/A, and C’A’ parallel to 
CA meeting CC’ in C’ and Pp in U. 

If then the hyperbola (a) is similar to the original hyperbola, it 
must by the last proposition be similar to the hyperbolic section 
made by the plane through C'’'A’U at right angles to the plane of 
the paper. 

Now C0O?-—CA"=(C'U*?—C'A”)+(CC'+0U)?— CC” 

> C'U?—C'A”, 
and PO .Op<PU. Up. 


lvili INTRODUCTION. 


Therefore PO. Op:CO?—CA"%<PU. Up: C'U?—C'A4 


and it follows that the hyperbolas are not similar*. 


4, Section of a spheroid by a plane parallel to the axis. 


That this is an ellipse similar to the generating ellipse can of 
course be proved in exactly the same way as theorem (2) above 
for the hyperboloid. 


* T think Archimedes is more likely to have used this proof than one on the 
lines suggested by Zeuthen (p. 421). The latter uses the equation of the 
hyperbola simply and proceeds thus. If y haye the same meaning as above, 
and if the coordinates of P referred to CA, CC’ as axes be 2, x, while those of O 
referred to the same axes are z, x’, we have, for the point P, 

x°=« (27—a?), 
where x is constant. 
Also, since the angle 4’C4 is given, «’=az, where a is constant. 


Thus Yea i} g?= (« be a?) 22 r καϑὶ 
Now z is proportional to CO, being in fact equal to Vice and the equation 
becomes 
᾿Ξ ΞΕ pat ᾿ 
YP Tat 6795 - καθ.......νυννννννννννννννννννον (1), 


which is clearly a hyperbola, since a2<k. 

Now, though the Greeks could have worked out the proof in a geometrical 
form equivalent to the above, I think that it is alien from the manner in which 
Archimedes regarded the equations to central conics, These he always expressed 
in the form of a proportion 


2 "2 2 
Ἢ a [ = = in the case of the ellipse | ᾽ 


wna? αἰ α 
and never in the form of an equation between areas like that used by 
Apollonius, viz. 
e—ps af τς 

Moreover the occurrence of the two different constants and the necessity 
of expressing them geometrically as ratios between areas and lines respectively 
would have made the proof very long and complicated ; and, as a matter of fact, 
Archimedes never does express the ratio y?/(x? -- a?) in the case of the hyperbola 
in the form of a ratio between constant areas like b?/a?. Lastly, when the 
equation of the given section through C'4’O was found in the form (1), assuming 
that the Greeks had actually found the geometrical equivalent, it would still 
have been held necessary, I think, to verify that 
ee) nar 

K-ay 

before it was finally pronounced that the hyperbola represented by the equation 
and the section made by the plane were one and the same thing. 


CAC 


RELATION OF ARCHIMEDES TO HIS PREDECESSORS. ΠΣ 


We are now in a position to consider the meaning of Archimedes’ 
remark that “the proofs of all these properties are manifest.” In 
the first place, it is not likely that “manifest” means “known” as 
having been proved by earlier geometers ; for Archimedes’ habit is 
to be precise in stating the fact whenever he uses important 
propositions due to his immediate predecessors, as witness his 
references to Eudoxus, to the Hlements [of Euclid], and to the 
“elements of conics.” When we consider the remark with reference 
to the cases of the sections parallel to the axes of the surfaces 
respectively, a natural interpretation of it is to suppose that 
Archimedes meant simply that the theorems are such as can easily 
be deduced from the fundamental properties of the three conics now 
expressed by their equations, coupled with the consideration that 
the sections by planes perpendicular to the axes are circles. But I 
think that this particular explanation of the ‘“‘ manifest” character 
of the proofs is not so applicable to the third of the theorems 
stating that any plane section of a hyperboloid of revolution 
through the vertex of the enveloping cone but not through the axis 
is a hyperbola. This fact is indeed no more “manifest” in the 
ordinary sense of the term than is the like theorem about the 
spheroid, viz. that any section through the centre but not through 
the axis is an ellipse. But this latter theorem is not given along 
with the other in Prop. 11 as being “manifest” ; the proof of it is 
included in the more general proposition (14) that any section of a 
spheroid not perpendicular to the axis is an ellipse, and that parallel 
sections are similar. Nor, seeing that the propositions are essen- 
tially similar in character, can I think it possible that Archimedes 
wished it to be understood, as Zeuthen suggests, that the proposition 
about the hyperboloid alone, and not the other, should be proved 
directly by means of the geometrical equivalent of the Cartesian 
equation of the conic, and not by means of the property of the 
rectangles under the segments of intersecting chords, used earlier 
[Prop. 3] with reference to the parabola and later for the case of 
the spheroid and the elliptic sections of the conoids and spheroids 
generally. This is the more unlikely, I think, because the proof 
by means of the equation of the conic alone would present much 
more difficulty to the Greek, and therefore could hardly be called 
“‘ manifest.” 

It seems necessary therefore to seek for another explanation, 
and I think it is the following. The theorems, numbered 1, 2, and 


lx INTRODUCTION. 


4 above, about sections of conoids and spheroids parallel to the axis 
are used afterwards in Props. 15—i7 relating to tangent planes ; 
whereas the theorem (3) about the section of the hyperboloid by a 
plane through the centre but not through the axis is not used in 
connexion with tangent planes, but only for formally proving that a 
straight line drawn from any point on a hyperboloid parallel to any 
transverse diameter of the hyperboloid falls, on the convex side of 
the surface, without it, and on the concave side within it. Hence 
it does not seem so probable that the four theorems were collected 
in Prop. 11 on account of the use made of them later, as that they 
were inserted in the particular place with special reference to the 
three propositions (12—14) immediately following and treating of the 
elliptic sections of the three surfaces. The main object of the whole 
treatise was the determination of the volumes of segments of the 
three solids cut off by planes, and hence it was first necessary to 
determine all the sections which were ellipses or circles and therefore 
could form the bases of the segments. Thus in Props. 12-14 
Archimedes addresses himself to finding the elliptic sections, but, 
before he does this, he gives the theorems grouped in Prop. 11 by 
way of clearing the ground, so as to enable the propositions about 
elliptic sections to be enunciated with the utmost precision. Prop. 
11 contains, in fact, explanations directed to defining the scope of 
the three following propositions rather than theorems definitely 
enunciated for their own sake; Archimedes thinks it necessary to 
explain, before passing to elliptic sections, that sections perpen- 
dicular to the axis of each surface are not ellipses but circles, and 
that some sections of each of the two conoids are neither ellipses nor 
circles, but parabolas and hyperbolas respectively. It is as if he had 
said, ‘‘ My object being to find the volumes of segments of the three 
solids cut off by circular or elliptic sections, I proceed to consider 
the various elliptic sections ; but I should first explain that sections 
at right angles to the axis are not ellipses but circles, while sections 
of the conoids by planes drawn in a certain manner are neither 
ellipses nor circles, but parabolas and hyperbolas respectively. With 
these last sections I am not concerned in the next propositions, and 
I need not therefore cumber my book with the proofs ; but, as some 
of them can be easily supplied by the help of the ordinary properties 
of conics, and others by means of the methods illustrated in the 
propositions now about to be given, I leave them as an exercise for 
the reader.” This will, I think, completely explain the assumption 


RELATION OF ARCHIMEDES TO HIS PREDECESSORS. Ixi 


of all the theorems except that concerning the sections of a spheroid 
parallel to the axis; and I think this is mentioned along with the 
others for symmetry, and because it can be proved in the same way 
as the corresponding one for the hyperboloid, whereas, if mention of 
it had been postponed till Prop. 14 about the elliptic sections of a 
spheroid generally, it would still require a proposition for itself, since 
the axes of the sections dealt with in Prop. 14 make an angle with 
the axis of the spheroid and are not parallel to it. 

At the same time the fact that Archimedes omits the proofs of 
the theorems about sections of conoids and spheroids parallel to the 
axis as “manifest” is in itself sufficient to raise the presumption 
that contemporary geometers were familiar with the idea of three 
dimensions and knew how to apply it in practice. This is no matter 
for surprise, seeing that we find Archytas, in his solution of the 
problem of the two mean proportionals, using the intersection of a 
certain cone with a curve of double curvature traced on a right 
circular cylinder*. But, when we look for other instances of early 
investigations in geometry of three dimensions, we find practically 
nothing except a few vague indications as to the contents of a lost 
treatise of Euclid’s consisting of two Books entitled Swrface-loci 
(τόποι πρὸς ἐπιφανείᾳ). This treatise is mentioned by Pappus 
among other works by Aristaeus, Euclid and Apollonius grouped 
as forming the so-called τόπος ἀναλυόμενος. As the other works in 
the list which were on plane subjects dealt only with straight lines, 
circles and conic sections, it is a prior? likely that the swrface-loci of 


* Cf. Eutocius on Archimedes (Vol. 11. pp. 98—102), or Apollonius of Perga, 
pp. Xxii.—xxilii. 

+ By this term we conclude that the Greeks meant ‘‘loci which are surfaces ” 
as distinct from loci which are lines. Cf. Proclus’ definition of a locus as 
‘*a position of a line or a surface involving one and the same property” 
(γραμμῆς ἢ ἐπιφανείας θέσις ποιοῦσα ἕν καὶ ταὐτὸν σύμπτωμα), p. 394. Pappus 
(pp. 660—2) gives, quoting from the Plane Loci of Apollonius, a classification of 
loci according to their order in relation to that of which they are the loci. Thus, 
he says, loci are (1) ἐφεκτικοί, i.e. fixed, e.g. in this sense the locus of a point is 
a point, of a linea line, and so on; (2) διεξοδικοί or moving along, a line being in 
this sense the locus of a point, a surface of a line, and a solid of a surface; 
(3) ἀναστροφικοί, turning backwards, i.e., presumably, moving backwards and 
forwards, a surface being in this sense the locus of a point, and a solid of a line. 
Thus a surface-locus might apparently be either the locus of a point or the 
locus of a line moving in space. 

+ Pappus, pp. 634, 636. 


lxil INTRODUCTION. 


Euclid included at least such loci as were cones, cylinders and 
spheres. Beyond this, all is conjecture based upon two lemmas 
given by Pappus in connexion with the treatise. 

First lemma to the Surface-loci of Huclid*. 

The text of this lemma and the attached figure are not satisfac- 
tory as they stand, but they have been explained by Tannery in a 
way which requires a change in the figure, but only the very slightest 
alteration in the text, as follows;. 

“Tf AB be a straight line and CD be parallel to a straight line 
given in position, and if the ratio AD. DB: DC’ be [given], the 
point C lies on a conic section. 
If now AB be no longer given in 
position and A, B be no longer 
given but lie on straight lines 
AE, EB given in positiont, the 
point C raised above [the plane, B 
containing AH, HB] is on a 
surface given in position. And 
this was proved.” 

According to this interpretation, it is asserted that, if A moves 
with one extremity on each of the lines AH, HB which are fixed, 
while DC is in a fixed direction and AD. DB: 2653 is constant, 
then Οὐ lies on a certain surface. So far as the first sentence is 
concerned, A remains of constant length, but it is not made 
precisely clear whether, when AB is no longer given in position, its 
length may also vary§. If however AB remains of constant length 
for all positions which it assumes, the surface which is the locus of 
C would be a complicated one which we cannot suppose that Euclid 
could have profitably investigated. It may, therefore, be that 
Pappus purposely left the enunciation somewhat vague in order to 
make it appear to cover several surface-loci which, though belonging 
to the same type, were separately discussed by Euclid as involving 


E 


* Pappus, p. 1004. 

+ Bulletin des sciences math., 2° Série, v1. 149. 

t+ The words of the Greek text are γένηται δὲ πρὸς θέσει εὐθεῖα ταῖς AE, EB, 
and the above translation only requires εὐθείαις instead of εὐθεῖα. The figure in 
the text is so drawn that ADB, AEB are represented as two parallel lines, and 
CD is represented as perpendicular to ADB and meeting AEB in E. 

§ The words are simply “if AB be deprived of its position (στερηθῇ τῆς 
θέσεως) and the points 4, B be deprived of their [character of] being given” 
(στερηθῇ τοῦ δοθέντος εἶναι). 


RELATION OF ARCHIMEDES TO HIS PREDECESSORS. [ΧΠῚ 


in each case somewhat different sets of conditions limiting the 
generality of the theorem. 

It is at least open to conjecture, as Zeuthen has pointed out*, 
that two cases of the type were considered by Euclid, namely, (1) 
that in which AB remains of constant length while the two fixed 
straight lines on which A, B respectively move are parallel instead 
of meeting in a point, and (2) that in which the two fixed straight 
lines meet in a point while A moves always parallel to itself 
and varies in length accordingly. 

(1) In the first case, where the length of AB is constant and 
the two fixed lines parallel, we should have a surface described by a 
conic moving bodily+. This surface would be a cylindrical surface, 
though it would only have been called a “ cylinder” by the ancients 
in the case where the moving conic was an ellipse, since the essence 
of a “cylinder” was that it could be bounded between two parallel 
circular sections. If then the moving conic was an ellipse, it would 
not be difficult to find the circular sections of the cylinder ; this 
could be done by first taking a section at right angles to the axis, 
after which it could be proved, after the manner of Archimedes, 
On Conoids and Spheroids, Prop. 9, first that the section is an ellipse 
or a circle, and then, in the former case, that a section made by 
a plane drawn at a certain inclination to the ellipse and passing 
through, or parallel to, the major axis is a circle. There was 
nothing to prevent Euclid from investigating the surface similarly 
generated by a moving hyperbola or parabola; but there would 
be no circular sections, and hence the surfaces might perhaps not 
have been considered as of very great importance, 

(2) In the second case, where AH, GH meet at a point and 
AB moves always parallel to itself, the surface generated is of 
course a cone. Some particular cases of this sort may easily have 
been discussed by Euclid, but he could hardly have dealt with the 
general case, where DC has any direction whatever, up to the 
point of showing that the surface was really a cone in the sense 
in which the Greeks understood the term, or (in other words) 
of finding the circular sections. To do this it would have been 
necessary to determine the principal planes, or to solve the dis- 


* Zeuthen, Die Lehre von den Kegelschnitten, pp. 425 sqq. 
+ This would give a surface generated by a moving line, διεξοδικὸς γραμμῆς 
as Pappus has it. 


lxiv INTRODUCTION. 


criminating cubic, which we cannot suppose Euclid to have done. 
Moreover, if Euclid had found the circular sections in the most 
general case, Archimedes would simply have referred to the fact 
instead of setting himself to do the same thing in the particular 
case where the plane of symmetry is given. These remarks apply 
to the case where the conic which is the locus of C is an ellipse ; 
there is still less ground for supposing that Euclid could have 
proved the existence of circular sections where the conic was a 
hyperbola, for there is no evidence that Euclid even knew that 
hyperbolas and parabolas could be obtained by cutting an oblique 
circular cone. 


Second lemma to the Surface-loci. 


In this Pappus states, and gives a complete proof of the propo- 
sition, that the locus of a point whose distance from a given point 
is in ὦ given ratio to its distance from a fixed line is a conic 
section, which is un ellipse, a parabola, or a hyperbola according 
as the given ratio is less than, equal to, or greater than unity*. 
Two conjectures are possible as to the application of this theorem 
by Euclid in the treatise referred to. 


(1) Consider a plane and a straight line meeting it at any angle. 
Imagine any plane drawn at right angles to the straight line and 
meeting the first plane in another straight line which we will call 
X. If then the given straight line meets the plane at right angles 
to it in the point S, a conic can be described in that plane with 
S for focus and X for directrix ; and, as the perpendicular on α΄ 
from any point on the conic is in a constant ratio to the per- 
pendicular from the same point on the original plane, all points 
on the conic have the property that their distances from S are in 
a given ratio to their distances from the given plane respectively. 
Similarly, by taking planes cutting the given straight line at right 
angles in any number of other points besides S, we see that the locus 
of a point whose distance from a given straight line is in a given 
ratio to its distance from a given plane is a cone whose vertex is 
the point in which the given line meets the given plane, while the 
plane of symmetry passes through the given line and is at right 
angles to the given plane. If the given ratio was such that the 
guiding conic was an ellipse, the circular sections of the surface 


* See Pappus, pp. 1006—1014, and Hultsch’s Appendix, pp. 1270—1273 ; or 
cf, Apollonius of Perga, pp. Xxxvi.—xxxviii, 


RELATION OF ARCHIMEDES TO HIS PREDECESSORS. lxv 


could, in that case at least, be found by the same method as 
that used by Archimedes (On Conoids and Spheroids, Prop. 8) in 
the rather more general case where the perpendicular from the 
vertex of the cone on the plane of the given elliptic section does 
not necessarily pass through the focus. 


(2) Another natural conjecture would be to suppose that, by 
means of the proposition given by Pappus, Euclid found the locus 
of a point whose distance from a given point is in a given ratio 
to its distance from a fixed plane. This would have given surfaces 
identical with the conoids and spheroids discussed by Archimedes 
excluding the spheroid generated by the revolution of an ellipse 
about the minor axis. We are thus brought to the same point as 
Chasles who conjectured that the Surfuce-loci of Euclid dealt with 
surfaces of revolution of the second degree and sections of the 
same*, Recent writers have generally regarded this theory as 
improbable. Thus Heiberg says that the conoids and spheroids 
were without any doubt discovered by Archimedes himself; other- 
wise he would not have held it necessary to give exact definitions 
of them in his introductory letter to Dositheus ; hence they could 
not have been the subject of Euclid’s treatiset. I confess I think 
that the argument of Heiberg, so far from being conclusive against 
the probability of Chasles’ conjecture, is not of any great weight. 
To suppose that Euclid found, by means of the theorem enunciated 
and proved by Pappus, the locus of a point whose distance from 
a given point is in a given ratio to its distance from a fixed plane 
does not oblige us to assume either that he gave a name to the 
loci or that he investigated them further than to show that sections 
through the perpendicular from the given point on the given plane 
were conics, while sections at right angles to the same perpendicular 
were circles ; and of course these facts would readily suggest them- 
selves. Seeing however that the object of Archimedes was to 
find the volumes of segments of each surface, it is not surprising 
that he should have preferred to give a definition of them which 
would indicate their form more directly than a description of them 
as loci would have done; and we have a parallel case in the dis- 
tinction drawn between conics as such and conics regarded as loci, 
which is illustrated by the different titles of Euclid’s Conics and 
the Solid Loci of Aristaeus, and also by the fact that Apollonius, 

* Apercu historique, pp. 273, 4. 
+ Litterargeschichtliche Studien tiber Euklid, Ὁ. 79. 


Ixvi_ INTRODUCTION. 


though he speaks in his preface of some of the theorems in his 
Conics as useful for the synthesis of ‘solid loci’ and goes on to 
mention the ‘locus with respect to three or four lines,’ yet enun- 
ciates no proposition stating that the locus of such and such a point 
is a conic. There was a further special reason for defining the 
conoids and spheroids as surfaces described by the revolution of 
a conic about its axis, namely that this definition enabled Archi- 
medes to include the spheroid which he calls ‘flat’ (ἐπιπλατὺ 
σφαιροειδές), 1.6. the spheroid described by the revolution of an 
ellipse about its minor axis, which is not one of the loci which 
the hypothesis assumes Euclid to have discovered. Archimedes’ 
new definition had the incidental effect of making the nature of 
the sections through and perpendicular to the axis of revolution 
even more obvious than it would be from Euclid’s supposed way 
of treating the surfaces; and this would account for Archimedes’ 
omission to state that the two classes of sections had been known 
before, for there would have been no point in attributing to Euclid 
the proof of propositions which, with the new definition of the 
surfaces, became self-evident. The further definitions given by 
Archimedes may be explained on the same principle. Thus the 
axis, as defined by him, has special reference to his definition of 
the surfaces, since it means the axis of revolution, whereas the 
axis of a conic is for Archimedes a diameter. The enveloping cone 
of the hyperboloid, which is generated by the revolution of the 
asymptotes about the axis, and the centre regarded as the point 
of intersection of the asymptotes were useful to Archimedes’ dis- 
cussion of the surfaces, but need not have been brought into 
Euclid’s description of the surfaces as loci. Similarly with the 
axis and vertex of a segment of each surface. And, generally, it 
seems to me that all the definitions given by Archimedes can be 
explained in like manner without prejudice to the supposed dis- 
covery of three of the surfaces by Euclid. 

I think, then, that we may still regard it as possible that 
Euclid’s Swrface-loct was concerned, not only with cones, cylinders 
and (probably) spheres, but also (to a limited extent) with three 
other surfaces of revolution of the second degree, viz. the paraboloid, 
the hyperboloid and the prolate spheroid. Unfortunately however 
we are confined to the statement of pussibilities; and certainty 
can hardly be attained unless as the result of the discovery of 
fresh documents. 


RELATION OF ARCHIMEDES TO HIS PREDECESSORS. lxvil 


§ 5. Two mean proportionals in continued proportion. 


Archimedes assumes the construction of two mean proportionals 
in two propositions (On the Sphere and Cylinder u. 1, 5). Perhaps 
he was content to use the constructions given by Archytas, 
Menaechmus*, and Eudoxus. It is worth noting, however, that 
Archimedes does not introduce the two geometric means where 
they are merely convenient but not necessary ; thus, when (On the. 


1 


Sphere and Cylinder τ. 34) he has to substitute for a ratio (f) : 
δ 
where β: γγ, a ratio between lines, and it is sufficient for his 
purpose that the required ratio cannot be greater than (Fy but 
7 


may be less, he takes two arithmetic means between β, y, as ὃ, ε, 
and then assumesy as a known result that 
ΒΕ: 
ey" 
* The constructions of Archytas and Menaechmus are given by Eutocius 
[Archimedes, Vol. 111. pp. 92—102]; or see Apollonius of Perga, pp. xix—xxiil. 


+ The proposition is proved by Eutocius; see the note to On the Sphere 
and Cylinder τ. 34 (p, 42). 


62 


CHAPTER IV. 


ARITHMETIC IN ARCHIMEDES. 


Two of the treatises, the Measurement of a circle and the 
Sand-reckoner, are mostly arithmetical in content. Of the Sand- 
reckoner nothing need be said here, because the system for expressing 
numbers of any magnitude which it unfolds and applies cannot be 
better described than in the book itself; in the Weaswrement of a 
circle, however, which involves a great deal of manipulation of 
numbers of considerable size though expressible by means of the 
ordinary Greek notation for numerals, Archimedes merely gives the 
results of the various arithmetical operations, multiplication, extrac- 
tion of the square root, etc., without setting out any of the operations 
themselves. Various interesting questions are accordingly involved, 
and, for the convenience of the reader, I shall first give a short 
account of the Greek system of numerals and of the methods by 
which other Greek mathematicians usually performed the various 
operations included under the general term λογιστική (the art of 
calculating), in order to lead up to an explanation (1) of the way in 
which Archimedes worked out approximations to the square roots of 
large numbers, (2) of his method of arriving at the two approximate 


values of V3 which he simply sets down without any hint as to how 
they were obtained*. 


* In writing this chapter I have been under particular obligations to Hultsch’s 
articles Arithmetica and Archimedes in Pauly-Wissowa’s Real-Encyclopiddie, τι. 
1, as well as to the same scholar’s articles (1) Die Néherungswerthe irrationaler 
Quadratwurzeln bei Archimedes in the Nachrichten von der kgl. Gesellschaft der 
Wissenschajten zu Gittingen (1893), pp. 367 sqq., and (2) Zur Kreismessung des 
Archimedes in the Zeitschrift fiir Math. u. Physik (Hist. litt. Abtheilung) xxxix. 
(1894), pp. 121 sqq. and 161 sqq. I have also made use, in the earlier part 
of the chapter, of Nesselmann’s work Die Algebra der Griechen and the histories 
of Cantor and Gow. 


ARITHMETIC IN ARCHIMEDES, lxix 


§ 1. Greek numeral system. 


It is well known that the Greeks expressed all numbers from 1 
to 999 by means of the letters of the alphabet reinforced by the 
addition of three other signs, according to the following scheme, in 
which however the accent on each letter might be replaced by a 
short horizontal stroke above it, as @. 

a, 8, y', ὃ, €,'s", ὁ ἢ, 0 are 1, 2, 3, 4, 5, 6, 7, 8, 9 respectively. 
MEU μον, δ, Oty G 55 LO, BOOK τῷ. 90 ες 
ΠΡΟ τὸ 6 OX, Ψ ὦ, 4 ,, 100, 200, 300,...... 900 95 

Intermediate numbers were expressed by simple juxtaposition 
(representing in this case addition), the largest number being placed 
on the left, the next largest following it, and so on in order. Thus 
the number 153 would be expressed by ρνγ' or pvy. There was no 
sign for zero, and therefore 780 was ψπ', and 306 rs" simply. . 

Thousands (χιλιάδες) were taken as units of a higher order, and 
1,000, 2,000, ... up to 9,000 (spoken of as χίλιοι, δισχίλιοι, κιτ.λ.} Were 
represented by the same letters as the first nine natural numbers 
but with a small dash in front and below the line; thus e.g. 6’ was 
4,000, and, on the same principle of juxtaposition as before, 1,823 was 
expressed by awxy’ or awxy, 1,007 by af’, and so on. 

Above 9,999 came a myriad (μυριάς), and 10,000 and higher 
numbers were expressed by using the ordinary numerals with the 
substantive μυριάδες taken as a new denomination (though the words 
μύριοι, δισμύριοι, τρισμύριοι, κιτ.λ. are also found, following the 
analogy of χίλιοι, δισχίλιοι and so on). Various abbreviations were 
used for the word μυριάς, the most common being M or Mv; and, 
where this was used, the number of myriads, or the multiple of 
10,000, was generally written over the abbreviation, though some- 


times before it and even after it. Thus 349,450 was Μθυν' ὩΣ 
Fractions (λεπτά) were written in a variety of ways. The most 
usual was to express the denominator by the ordinary numeral with 
two accents affixed. When the numerator was unity, and it was 
therefore simply a question of a symbol for a single word such as 
* Diophantus denoted myriads followed by thousands by the ordinary signs 
for numbers of units, only separating them by a dot from the thousands. Thus 


for 3,069,000 he writes rs. 0, and λγ. ἀψος for 331,776. Sometimes myriads 
were represented by the ordinary letters with two dots above, as p =100 myriads 
(1,000,000), and myriads of myriads with two pairs of dots, as ¢ for 10 myriad- 
myriads (1,000,000,000). 


Ixx INTRODUCTION. 


τρίτον, 1, there was no need to express the numerator, and the 


symbol was γ΄; similarly ς΄ -- 1, ιε΄ -- 4,, and so on. When the 


numerator was not unity and a certain number of fourths, fifths, 
c., had to be expressed, the ordinary numeral was used for the 


numerator; thus θ΄ ια΄ = %, ι΄ οα΄ --19. In Heron’s Geometry the 
denominator was written twice in the latter class of fractions ; thus 


ents, 23 


δύο πέμπτα) was β' εἴ ε $3 (λεπτὰ τ ιακοστότριτα κγ΄ ΟΥ εἰκοσιτρία 
μ ᾽ ρ ρ γ ρ 


σαι 


τριακοστότριτα) Was Ky’ dy” λγ΄. The sign for 3, ἥμισυ, is in 
Archimedes, Diophantus and Eutocius _”, in Heron C or a sign 
similar to a capital S*. 

A favourite way of expressing fractions with numerators greater 
than unity was to separate them into component fractions with 
numerator unity, when juxtaposition as usual meant addition. Thus 
# was written ἰ.. δ΄ -  Ε1, 18 was Cd’yis”=4+44+44+3; 
Eutocius writes £6” or $+, for 23, and so on. Sometimes the 
same fraction was separated into several different sums; thus in 
Heron (p. 119, ed. ey 363 is variously expressed as 


224 


(0) $+7+30+a:2+2Rn 
(6) 3+3+ fan ay 
and (0) ἐ Ἐπ τ οὖν Ἐτῖσ Ἐ αὖτ. 


Seaagesimal fractions. This system has to be mentioned because 
the only instances of the working out of some arithmetical operations 
which have been handed down to us are calculations expressed in 
terms of such fractions; and moreover they are of special interest 
as having much in common with the modern system of decimal 
fractions, with the difference of course that the submultiple is 60 
instead of 10. The scheme of sexagesimal fractions was used by the 
Greeks in astronomical calculations and appears fully developed in 
the σύνταξις of Ptolemy. The circumference of a circle, and along 
with it the four right angles subtended by it at the centre, are 
divided into 360 parts (τμήματα or μοῖραι) or as we should say degrees, 
each μοῖρα into 60 parts called (first) sixtieths, (πρῶτα) ἑξηκοστά, 
or minutes (λεπτά), each of these again into δεύτερα ἑξηκοστά (seconds), 
and so on. <A similar division of the radius of the circle into 60 

* Diophantus has a general method of expressing fractions which is the 


exact reverse of modern practice; the denominator is written above the 
Ύ κε a. wis 


numerator, thus ¢=5/3, κα = 21/25, and pxf. φξη =1,270,568/10,816. Some- 
times he writes down the numerator and then introduces the denominator 


with ἐν μορίῳ or μορίου, e.g. Ts .,8 wop. Ny. αψος = 3,069,000/331,776. 


ARITHMETIC IN ARCHIMEDES. xxi 


parts (τμήματα) was also made, and these were each subdivided into 
sixtieths, and so on. Thus a convenient fractional system was 
available for general arithmetical calculations, expressed in units of 
any magnitude or character, so many of the fractions which we 
should represent by εἶσ» so many of those which we should write 
(25), (gs), and so on to any extent. It is therefore not surprising 
that Ptolemy should say in one place “In general we shall use the 
method of numbers according to the sexagesimal manner because of 
the inconvenience of the [ordinary] fractions.” For it is clear that 
the successive submultiples by 60 formed a sort of frame with fixed 
compartments into which any fractions whatever could be located, 
and it is easy to see that e.g. in additions and subtractions the 
sexagesimal fractions were almost as easy to work with as decimals 
are now, 60 units of one denomination being equal to one unit of 
the next higher denomination, and “carrying” and “borrowing” 
being no less simple than it is when the number of units of one 
denomination necessary to make one of the next higher is 10 instead 
of 60. In expressing the units of the circumference, degrees, μοῖραι 
or the symbol 4 was generally used along with the ordinary numeral 
which had a stroke above it ; minutes, seconds, etc. were expressed 
by one, two, etc. accents affixed to the numerals. Thus fi B=2°, 
μοιρῶν pl μβ΄ μ' =47° 42’ 40". Also where there was no unit in any 
particular denomination O was used, signifying οὐδεμία μοῖρα, οὐδὲν 
ἑξηκοστόν and the like; thus Oa’ β΄ O’’=0° 1΄ 2" 0". Similarly, for 
the units representing the divisions of the radius the word τμήματα 
or some equivalent was used, and the fractions were represented as 


before ; thus τμημάτων ἕζ δ' ve” = 67 (units) 4’ 55”. 
§ 2. Addition and Subtraction. 


There is no doubt that, in writing down numbers for these 
purposes, the several powers of 10 were kept separate in a manner 
corresponding practically to our system of numerals, and the 
hundreds, thousands, etc., were written in separate vertical rows, 
The following would therefore be a typical form of a sum in addition ; 

αυκδ' = 1424 
pon 103 


Mora’ 12281 
M 30030 
ὃ 

M yor η΄ 43838 


Ixxil INTRODUCTION. 


and the mental part of the work would be the same for the Greek as 
for us. 


Similarly a subtraction would be represented as follows: 
θ 
M yxAs’ = 93636 
B 
My θ΄ 23409 


M σκζ Τ0227 

§ 8. Multiplication. 

A number of instances are given in Eutocius’ commentary on 
the Measurement of a circle, and the similarity to our procedure is 
just as marked as in the above cases of addition and subtraction. 
The multiplicand is written first, and below it the multiplier preceded 
by ἐπί (=“‘into”). Then the highest power of 10 in the multiplier 
is taken and multiplied into the terms containing the separate 
multiples of the successive powers of 10, beginning with the highest 
and descending to the lowest ; after which the next highest power 
of 10 in the multiplier is multiplied into the various denominations 
in the multiplicand in the same order. The same procedure is 
followed where either or both of the numbers to be multiplied 
contain fractions. Two instances from Eutocius are appended from 
which the whole procedure will be understood. 


(1) ie 780 
ἐπὶ wr’ x 780 
Oe 
MM ς΄ 490000 56000 
Mosul 56000 6400 
é 
ὁμοῦ M nv’ sum 608400 
(2) 
By 8? 30133 } [=30133] 
ἐπὶ γιγ' "8" x 30133 4 
ει Χ 
MM 6 adyv’ 9,000,000 30,000 9,000 1500 750 
Μρλέβ' 1." 30,000 100 ὃ. Ὁ - 
Dre ea igh 9,000 30 9 ὙΠ 
αφ' εα' ἔδυ 1,500 5 11 1 = 
yr’ β΄ hee Eo sy 750 24 4 ἊΣ 1 ἑ τῷ 


Ἂ 
[ὁμοῦ] Μβχαθιις" [9,041,250 + 30,1374 + 9,041} + 1506+4+}42 
+7534+4434+ 
= 9,082,689,),. 


ARITHMETIC IN ARCHIMEDES. Ixxili 


One instance of a similar multiplication of numbers involving 
fractions may be given from Heron (pp. 80, 81). It is only one of 
many, and, for brevity, the Greek notation will be omitted. Heron 
has to find the product of 423 and 782, and proceeds as follows: 


64) 
1.7 58 
Δ 5 
4. ee ΞΞ πες 


9) 510 62 H lng ene 62 6 1 
28 +e t+ ee ec = 28 ἢ Ὁ 51 ot tz 
5+ a 


The multiplication of 37° 4’ 55” (in the sexagesimal system) by 
itself is performed by Theon of Alexandria in his commentary on 
Ptolemy’s σύνταξις in an exactly similar manner. 


§ 4. Division. 


The operation of dividing by a number of one digit only was 
easy for the Greeks as for us, and what we call “long division” was 
with them performed, mutatis mutandis, in the same way as now 
with the help of multiplication and subtraction. Suppose, for 
instance, that the operation in the first case of multiplication given 


above had to be reversed and that Μην (608,400) had to be divided 
by Wz’ (780). The terms involving the different powers of 10 would 
be mentally kept separate as in addition and subtraction, and the 
first question would be, how many times will 7 hundreds go into 60 
myriads, due allowance being made for the fact that the 7 hundreds 
have 80 behind them and that 780 is not far short of 8 hundreds ? 
The answer is 7 hundreds or y’, and this multiplied by the divisor 


vd ξ 
Wx’ (780) would give Ms’ (546,000) which, subtracted from M ηυ' 


(608,400), leaves the remainder M Bu’ (62,400). This remainder has 
then to be divided by 780 or a number approaching 8 hundreds, and 
8 tens or π' would have to be tried. In the particular case the 
result would then be complete, the quotient being ψπ' (780), and 
there being no remainder, since π΄ (80) multiplied by wz’ (780) gives 


the exact figure M βυ' (62,400). 


lxxiv INTRODUCTION. 


An actual case of long division where the dividend and divisor 
contain sexagesimal fractions is described by Theon. The problem 
is to divide 1515 20’ 15” by 25 12’ 10”, and Theon’s account of the 
process comes to this. 


Divisor Dividend Quotient 
25-12) 10? 515 20% 15” First term 60 
a 25. 60 = 1500 

Remainder 15 = 900’ 
Sum πο 
127160 — 720’ 
Remainder 200’ 
10%. Cor 10’ 
Remainder 190’ Second term 7’ 
ae (= 175. 
τ 15’ = 900 
Sum Oth” 
ibe ay fe 84” 
Remainder Sole 
LO 1.7. Oe 
Remainder 829” 50” |Third term 33” 
2,7) 1 ΘΠ, 825" 
Remainder 4" 50" = 290” 
19 1955 996" 


(too great by) 106” 


Thus the quotient is something less than 60 7’ 33”. It will be 
observed that the difference between this operation of Theon’s and 


that followed in dividing Mee (608,400) by ψπ' (780) as above is 
that Theon makes three subtractions for one term of the quotient, 
whereas the remainder was arrived at in the other case after one 
subtraction. The result is that, though Theon’s method is quite 
clear, it is longer, and moreover makes it less easy to foresee what 
will be the proper figure to try in the quotient, so that more time 
would be apt to be lost in making unsuccessful trials. 


§ 5. Extraction of the square root. 


We are now in a position to see how the operation of extracting 
the square root would be likely to be attacked, First, as in the case 
of division, the given whole number whose square root is required 
would be separated, so to speak, into compartments each containing 


ARITHMETIC IN ARCHIMEDES. Ixxv 


such and such a number of units and of the separate powers of 10. 
Thus there would be so many units, so many tens, so many hundreds, 
etc., and it would have to be borne in mind that the squares of 
numbers from 1 to 9 would lie between 1 and 99, the squares of 
numbers from 10 to 90 between 100 and 9900, and so on. Then the 
first term of the square root would be some number of tens or 
hundreds or thousands, and so on, and would have to be found in 
much the same way as the first term of a quotient in a “long 
division,” by trial if necessary. If A is the number whose square 
root is required, while a represents the first term or denomination of 
the square root and « the next term or denomination still to be 
found, it would be necessary to use the identity (a + #)?=a? + 2aa +a 
and to find « so that 2ax+a? might be somewhat less than the 
remainder 4—a?, Thus by trial the highest possible value of a 
satisfying the condition would be easily found. If that value were 
ὁ, the further quantity 2ab +6? would have to be subtracted from 
the first remainder A — a’, and from the second remainder thus left 
a third term or denomination of the square root would have to be 
derived, and so on. That this was the actual procedure adopted is 
clear from a simple case given by Theon in his commentary on the 
σύνταξις. Here the square root of 144 is in question, and it is 
obtained by means of Eucl. 1. 4. The highest possible denomina- 
tion (i.e. power of 10) in the square root is 10 ; 10? subtracted from 
144 leaves 44, and this must contain not only twice the product of 
10 and the next term of the square root but also the square of that 
next term itself. Now, since 2.10 itself produces 20, the division 
of 44 by 20 suggests 2 as the next term of the square root; and 
this turns out to be the exact figure required, since 


2. 20+ 2° = 44. 


The same procedure is illustrated by Theon’s explanation of 
Ptolemy’s method of extracting square roots according to the 
sexagesimal system of fractions. The problem is to find approxi- 
mately the square root of 4500 μοῖραι or degrees, and a geometrical 
figure is used which makes clear the essentially Euclidean basis of 
the whole method. Nesselmann gives a complete reproduction of 
the passage of Theon, but the following purely arithmetical represen- 
tation of its purport will probably be found clearer, when looked at 
side by side with the figure. 


Ptolemy has first found the integral part of /4500 to be 67. 


lxxvl INTRODUCTION. 


Now 67? = 4489, so that the remainder is 11. Suppose now that 
the rest of the square root is expressed by means of the usual 
sexagesimal fractions, and that we may therefore put 


45002 67? s LL Sen ee ποῦ 


60 * 60?’ 
2.67x 
where a, y are yet to be found. Thus ὦ must be such that 60 
: 11.60 
is somewhat less than 11, or x must be somewhat less than 267 


or which is at the same time greater than 4. On trial, it 


oT 
turns out that 4 will satisfy the conditions of the problem, namely 


2 


that (67 + a) must be less than 4500, so that a remainder will 


be left by means of which y may be found. 


a n K ὃ 
| 67° 4! 55” 
4489 268’ | & 
= 
a 
| 3 
| of 
εἶ - 
§ 
4’ 268' 16” 
g λ 
55” 3688” 40" | 
β Y 
2.67.4 AN? ; ΔΝ 
Now 11 — ————_ — & is the remainder, and this is equal to 
60 60 
11; 60'—2.67.4560—16 Ὁ 7155 
60° ἡ oo 
4\ y : 1424 
᾿ : G7 acess baie - 
Thus we must suppose that 2 (67 + a) 602 approximates to 60F ? 


or that 8048y is approximately equal to 7424. 60. 


ARITHMETIC IN ARCHIMEDES. ΙΧΧΥΙ 


Therefore y is approximately equal to 55. We have then to 
subtract 


: Ὁ ΕΣ BB\?  Ἀ442640 805 
2( 60) 60? τε. oF 60° * 60F” 


: 7424 
from the remainder 0" above found. 


442640 1424. 2800 46 40 


The subtraction of $0? from 602 δ “Gos > Of Boe δ’ 
3025 


OS 


but Theon does not go further and subtract the remaining 


ἼΩΝ 
60? 
As a matter of fact, if we deduct the 


instead of which he merely remarks that the square of 


40 
60?” 60%" 
3025 £ 2800 : ; ΐ bee 
Gor [πὰ gor» 80 as to obtain the correct remainder, it is 
164975 

Coe” 
To show the power of this method of extracting square roots by 
means of sexagesimal fractions, it is only necessary to mention that 

δ 28 
60 60) 60° 
approximation is equivalent to 1:7320509 in the ordinary decimal 
notation and is therefore correct to 6 places. 

But it is now time to pass to the question how Archimedes 


approximates to 


found to be 


Ptolemy gives as an approximation to \/3, which 


obtained the two approximations to the value of /3 which he 
assumes in the Measwrement of a circle. In dealing with this 
subject I shall follow the historical method of explanation adopted 
by Hultsch, in preference to any of the mostly ὦ priori theories 
which the ingenuity of a multitude of writers has devised at 
different times. 


§ 6. Early investigations of surds or incommensurables. 


From a passage in Proclus’ commentary on Eucl. 1.* we learn 
that it was Pythagoras who discovered the theory of irrationals 
(ἡ τῶν ἀλόγων πραγματεία). Further Plato says (Theaetetus 147 Ὁ), 
“On square roots this Theodorus [of Cyrene] wrote a work in 


* p. 65 (ed. Friedlein). 


lxxvili INTRODUCTION. 


which he proved to us, with reference to those of 3 or 5 [square] feet 
that they are incommensurable in length with the side of one square 
foot, and proceeded similarly to select, one by one, each [of the other 
incommensurable roots] as far as the root of 17 square feet, beyond 


which for some reason he did not go.” The reason why ν 3 is not 
mentioned as an incommensurable square root must be, as Cantor 
says, that it was before known to be such. We may therefore 
conclude that it was the square root of 2 which was geometrically 
constructed by Pythagoras and proved to be incommensurable with 
the side of a square in which it represented the diagonal. A clue 
to the method by which Pythagoras investigated the value of J2 
is found by Cantor and Hultsch in the famous passage of Plato 
(Rep. vit. 546 B, c) about the ‘geometrical’ or ‘nuptial’ number. 
Thus, when Plato contrasts the ῥητὴ and ἄρρητος διάμετρος τῆς 
πεμπάδος, he is referring to the diagonal of a square whose side 
contains five units of length ; the appyros διάμετρος, or the irrational 
diagonal, is then /50 itself, and the nearest rational number is 
J/50—1, which is the ῥητὴ διάμετρος. We have herein the 
explanation of the way in which Pythagoras must have made the 
first and most readily comprehensible approximation to /2; he 
must have taken, instead of 2, an improper fraction equal to it but 
such that the denominator was a square in any case, while the 
numerator was as near as possible to a complete square. Thus 


Pythagoras chose and the first approximation to /2 was 


50 


fas : - ; 
accordingly 5? it being moreover obvious that / I>. Again, 


Pythagoras cannot have been unaware of the truth of the 
proposition, proved in Eucl. τι. 4, that (a+ δ)" Ξ- αὐ + 2ab +b’, where 
a, b are any two straight lines, for this proposition depends solely 
upon propositions in Book 1. which precede the Pythagorean 
proposition 1. 47 and which, as the basis of 1. 47, must necessarily 
have been in substance known to its author. <A slightly different 
geometrical proof would give the formula (α -- ὁ)" -- αὖ -- 3αὖ - δ᾽, 
which must have been equally well known to Pythagoras. It could 
not therefore have escaped the discoverer of the first approximation 


J/50—1 for /50 that the use of the formula with the positive sign 


: ΕΠ : 1 ean 
would give a much nearer approximation, viz. 7 + ——, which is only 


14’ 


ARITHMETIC IN ARCHIMEDES. lxxix 


14 
assign to Pythagoras the discovery of the fact represented by 


greater than /50 to the extent of (=): Thus we may properly 


ron ἢ EN 
{πω 50> 7. 


The consequential result that /2> ; /50—1 is used by 


Aristarchus of Samos in the 7th proposition of his work On the 
size and distances of the sun and moon*. 


With reference to the investigations of the values of /3, V5, 


Gon. J17 by Theodorus, it is pretty certain that /3 was 
geometrically represented by him, in the same way as it appears 


* Part of the proof of this proposition was a sort of foretaste of the first part 
of Prop. 3 of Archimedes’ Measurement of a 
circle, and the substance of it is accordingly A 
appended as reproduced by Hultsch. 

ABEK is a square, KB a diagonal, 2 HBH 
=i4KBE, 2 FBE=3°,and AC is perpendicu- 
lar to BF so that the triangles ACB, BEF are 
similar. 

Aristarchus seeks to prove that 

Ald8Y & TeXOF se 1S} 2 1 

If R denote a right angle, the angles KBE, 
HBE, FBE are respectively 3910, 15}, 2,R. B = 

Then HE: FE > ZHBE: Z£FBE. 


[This is assumed as a known lemma by Aristarchus as well as Archimedes. ] 


K 


Therefore γε BD ag) 2) ee ee See ee (a). 
Now, by construction, BK?=2BE?, 
Also [Eucl. vr. 3] ΒΚ 2 ois HOPI SED) ® 

whence KH=N2HE. 


And, since N2 =x γ33Ξ: Ρ 
25 


HiGal G83) S75 15%, 
so that ἜΗΙ Ds Re) CEN | a Race (g). 
From (a) and (8), ex aequali, 
KE: FE > 18:1. 
Therefore, since BF > BE (or KE), 


BF: FE > 18: 1, 
so that, by similar triangles, 
ABT BC = Nis le 


Ixxx INTRODUCTION. 


afterwards in Archimedes, as the perpendicular from an angular 
point of an equilateral triangle on the opposite side. It would 
thus be readily comparable with the side of the “1 square foot” 
mentioned by Plato. The fact also that it is the side of three 
square feet (τρίπους δύναμις) which was proved to be incommensurable 
suggests that there was some special reason in Theodorus’ proof for 
specifying feet, instead of units of length simply; and the ex- 
planation is probably that Theodorus subdivided the sides of his 
triangles in the same way as the Greek foot was divided into 
halves, fourths, eighths and sixteenths. Presumably therefore, 


exactly as Pythagoras had approximated to /2 by putting = 


for 2, Theodorus started from the identity 3= Ἐξ It would then 


16 
= ἴεν, ἢ 
Mie ae Ae 


To investigate ./48 further, Theodorus would put it in the form 
/49—1, as Pythagoras put ./50 into the form J/49+1, and the 
result would be 


be clear that 


/48 (= eee 


We know of no further investigations into incommensurable 
square roots until we come to Archimedes. 


5.7. Archimedes’ approximations to V3. 


Seeing that Aristarchus of Samos was still content to use the 
first and very rough approximation to ν 2 discovered by Pythagoras, 
it is all the more astounding that Aristarchus’ younger contemporary 
Archimedes should all at once, without a word of explanation, give 
out that 

1351 - 265 
780 > V5 T53° 
as he does in the Measurement of a circle. 

In order to lead up to the explanation of the probable steps by 
which Archimedes obtained these approximations, Hultsch adopts 
the same method of analysis as was used by the Greek geometers in 
solving problems, the method, that is, of supposing the problem 
solved and following out the necessary consequences. ΤῸ compare 


ARITHMETIC IN ARCHIMEDES. ΙΧΧΧῚ 


265 1351 sey : ς 
the two fractions -- 153 ὃ and 7B” we first divide both denominators 


* into their smallest factors, and we obtain 
ΠΟ ΞΟ 229.52 139, 


fod =o. 1 
We observe also that 2.2.13 =52, while 3.17 =51, and we may 
therefore show the Ἔν between the numbers thus, 
(80'= 3.5. 2, 
Ξε οἷς 


For convenience of comparison we multiply the numerator and 


denominator of = by 5; the two original fractions are then 
1351 ἃ 1325 
15. Sou dae ol 

so that we can put Archimedes’ assumption in the form 


1351 1325 
ΠΡΌΣ ΙΝ: 


> 15/3 > 
and this is seen to be equivalent to 


ieee Die Oras 


1 : 
Now 26-5 = / 2 6°—1+ (ss) , and the latter expression 


is an approximation to /26?—1. 


We have then 26 = > /26? -- 1. = jl. 


As La was compared with 15,/3, and we want an ap- 


proximation to ./3 itself, we divide by 15 and so obtain 


ἡ (28- 5) > τὸ 28-1. 


15 52 
LO ere 676-1 Bi ΤΤς Ὁ ͵ 
Bae 15 V26?-1= \/ 995. ~ A/ 995. ~ /3, and it follows 


that - (26 ~55)> "5" 


The lower limit for /3 was given by 


1 1 


. xxxii INTRODUCTION. 


and a glance at this suggests that it may have been arrived at by 
simply substituting (52 -- 1) for 52. 

Now as a matter of fact the following proposition is true. Jf 
a’ +b is a whole number which is not a square, while a? is the nearest 
square number (above or below the first number, as the case may be), 
then 
b 


ὖ 
Cis -- Jato +b>at 
2a 2a+1° 


Hultsch proves this pair of inequalities in a series of propositions 
formulated after the Greek manner, and there can be little doubt 
that Archimedes had discovered and proved the same results in 
substance, if not in the same form. ‘The following circumstances 
confirm the probability of this assumption. 

(1) Certain approximations given by Heron show that he 
knew and frequently used the formula 


ρος b 
V@tbwats, 
«Ὁ 


(where the sign οὦ denotes ‘‘is approximately equal to”). 


Thus he gives /50 07 + = : 
Vopese= 
16’ 
11 
V75 οὐ 8 Ὁ 16° 
(2) The formula Ja*+boa+ ἘΠ i is used by the Arabian 
Alkarkhi (11th century) who drew from Greek sources (Cantor, 
Ῥ 719'sq.). 


It can therefore hardly be accidental that the formula 
b = b 
+ — a+b>at+-— 
Be ko eas = 9a41 
gives us what we want in order to obtain the two Archimedean 
approximations to ν 9, and that in direct connexion with one 
another*. 

* Most of the a priori theories as to the origin of the approximations are 
open to the serious objection that, as a rule, they give series of approximate 
values in which the two now in question do not follow consecutively, but are 
separated by others which do not appear in Archimedes. Hultsch’s explanation 


is much preferable as being free from this objection. But it is fair to say that 
the actual formula used by Hultsch appears in Hunrath’s solution of the puzzle 


ARITHMETIC IN ARCHIMEDES. lxxxili 


We are now in a position to work out the synthesis as follows. 


From the geometrical representation of /3 as the perpendicular 
from an angle of an equilateral triangle on the opposite side we 


obtain /2?—1= J3 and, as a first approximation, 


2 on : > 3. 
Using our formula we can transform this at once into 
= il 1 
2 —-_—. 2---. 
ν8-:- resins 3 


1 5 
Archimedes would then square (2 - 3) or 3 and would obtain 


; Le. he would put 


which he would compare with 3, or = : 


3 (5+ 5)> V3, ie. 2 > 0/8. 


3 
To obtain a still nearer approximation, he would proceed in the 
nee 26\? 676 h3, 675 h : 
same manner and compare (=) 1 OF 558» wit or 555,» Whence it 
5 Adel 
would appear that /3 = 995? 
Mites rore that 1. ὦ 
and therefore tha iB (2 -- 5) Say 
: 1351 

that is, 780 > J3. 


The application of the formula would then give the result 


5] 1 
=. (eer 
V3 > τὸ (26 a1) 


1326 —1 265 
Tosnie 7 + 153" 
The complete result would therefore be 


1351 aS 265 

780 153° 
(Die Berechnung irrationaler Quadratwurzeln vor der Herrschaft der Decimal- 
briiche, Kiel, 1884, p, 21; ef. Ueber das Ausziehen der Quadratwurzel bei 
Griechen und Indern, Hadersleben, 1883), and the same formula is implicitly 
used in one of the solutions suggested by Tannery (Sur la mesure du cercle 
ad Archiméde in Mémoires de la société des sciences physiques et naturelles de 
Bordeaux, 2° série, 1v. (1882), p. 313-337). 


that is, V3 > 


f2 


lxxxiv INTRODUCTION. 


Thus Archimedes probably passed from the first approximation 
a to from 1 to BBS and from = 


directly to the closest 


Lape 
+ 3) 3, a kp 5 780 ’ 
approximation of all, from which again he derived the less close 


imation 28} 
approxima 153° 
nearer approximation than | is probably that the squaring of 
this fraction would have brought in numbers much too large to be 


conveniently used in the rest of his calculations. A similar reason 


The reason why he did not proceed to a still 


‘will account for his having started from 5 instead of Ἷ ; if he had 


used the latter, he would first have obtained, by the same method, 


ea | - τ πς U lee ἢ a 
Ne 5° ane /3, or 567 /3; the squaring 
Oy pee 
of ea would have given ν ὃ-- oe , and the corresponding 
18817 


approximation would have given 56.194? where again the numbers 
are inconveniently large for his purpose. 

§ 8. Approximations to the square roots of large 
numbers. 


Archimedes gives in the Measurement of a circle the following 
approximate values : 


(1) 30132 > /9082321, 
(2) 18382, > 3380929, 
(3) 10093 > 1018405, 
(4) 20174 > /4069284,1,, 
(5) 5911 </349450, 

(6) 11721 < se 
(7) 23394 < /54721322., 


There is no doubt that in obtaining the integral portion 
of the square root of these numbers Archimedes used the method 
based on the Euclidean theorem (a+b)? = a?+ 2ab +6? which has 


ARITHMETIC IN ARCHIMEDES. Ixxxv 


already been exemplified in the instance given above from Theon, 
where an approximation to /4500 is found in sexagesimal fractions. 
The method does not substantially differ from that now followed; but 
whereas, to take the first case, 9082321, we can at once see what 
will be the number of digits in the square root by marking off pairs 
of digits in the given number, beginning from the end, the absence 
of a sign for Ὁ in Greek made the number of digits in the square 
root less easy to ascertain because, as written in Greek, the number 


Μ᾿ Brea! only contains six signs representing digits instead of seven. 
Even in the Greek notation however it would not be difficult to see 
that, of the denominations, units, tens, hundreds, etc. in the square 
root, the units would correspond to xa’ in the original number, the 


Ul Ἂ 
tens to Br, the hundreds to M, and the thousands to M. Thus it 
would be clear that the square root of 9082321 must be of the form 


1000a + 100y + 10z + w, 


where «, y, z, w can only have one or other of the values 0, 1, 2,...9. 
Supposing then that x is found, the remainder V —(1000z)’, where 
WV is the given number, must next contain 2.1000x”.100y and 
(100y)’, then 2(1000xz+100y).10z and (10z)’, after which the 
remainder must contain two more numbers similarly formed. 


In the particular case (1) clearly e=3, The subtraction of 
(3000)? leaves 82321, which must contain 2.3000.100y. But, even 
if y is as small as 1, this product would be 600,000, which is greater 
than 82321. Hence there is no digit representing hundreds in the 
square root. To find z, we know that 82321 must contain 


2.3000. 102 + (10z)’, 


and z has to be obtained by dividing 82321 by 60,000. Therefore 
z=1. Again, to find w, we know that the remainder 


(82321 — 2.3000. 10 -- 105), 


or 22221, must contain 2.3010w+w*, and dividing 22221 by 
2.3010 we see that w=3. Thus 3013 is the integral portion of 
the square root, and the remainder is 22221 —(2.3010.3+ 3°), or 
4152. 

The conditions of the proposition now require that the approxi- 
mate value to be taken for the square root must not be less than 


lxxxvi INTRODUCTION. 


the real value, and therefore the fractional part to be added to 3013 
must be if anything too great. Now it is easy to see that the 


2 
fraction to be added is greater than : because 2.3013. _ (5) is 


less than the remainder 4152. Suppose then that the number 
required (which is nearer to 3014 than to 3013) is 3014, 


and - has to be if anything too small. 


Now (3014)? = (3013)? + 2. 3013 + 1 = (3013)? + 6027 
= 9082321 -- 4152 + 6027, 


whence 9082321 = (3014)? -- 1875. 
By applying Archimedes’ formula Ja? +b<a+ 5 , we obtain 
3014 -- oe 9082321. 
The required value / Ξ has therefore to be not greater than δ ae : 
It remains to be explained why Archimedes put for δ the value 4 
which is equal to oe In the first place, he evidently preferred 


fractions with unity for numerator and some power of 2 for 
denominator because they contributed to ease in working, e.g. when 
two such fractions, being equal to each other, had to be added. 


The exceptions, the fractions me and τ are to be explained Ὁ 
" Tyas ᾿ ἦ 


exceptional circumstances presently to be mentioned.) Further, in 
the particular case, it must be remembered that in the subsequent 


work 2911 had to be added to 3014 —F and the sum divided by 780, 


or 2.2.3.5.13. It would obviously lead to simplification if a 

factor could be divided out, e.g. the best for the purpose, 13. Now, 

dividing 2911 + 3014, or 5925, by 13, we obtain the quotient 455, 
P 


and a remainder 10, so that ae remains to be divided by 13. 
Therefore ᾧ has to be so chosen that 10g -- Ὁ is divisible by 13, while 


i approximates to, but is not greater than, ee The solution 


p=1,q=4 would therefore be natural and easy. 


ARITHMETIC IN ARCHIMEDES. lxxxvll 


(2) »/3380929. 

The usual process for extraction of the square root gave as the 
integral part of it 1838, and as the remainder 2685. As before, it 
was easy to see that the exact root was nearer to 1839 than to 1838, 
and that 


/3380929 = 1838? + 2685 = 1839" -- 2. 1838 —1 + 2685 


= 1839? — 992. 
The Archimedean formula then gave 
992 ππαπασς 
sane 2 
1839 — στο. 9" V3380929. 


It could not have escaped Archimedes that : 


992 , 1984 : 1 1839 1 
3678 Of 73567 1508 G = 735° and Z would have satisfied 


the necessary condition that the fraction to be taken ἘΠΕ be less 


was a near approxima- 


tion to 


than the real value. Thus it is clear that, in taking es as the 


approximate value of the fraction, Archimedes had in view the 
simplification of the we. work by the elimination of a factor, 


If the fraction be denoted by Ὁ , the sum of 1839-7 and 1823, or 


3662 -- a had to be divided by en ie. by 6.40. Division of 3662 
by 40 gave 22 as remainder, and then », ῳ had to be so chosen that 


22-7 was conveniently divisible by 40, while 7 was less than but 


9 
approximately equal to ee The solution p= 2, g=11 was easily 


3678 ° 
seen to satisfy the conditions. 


(3) »/1018405. 


The usual procedure gave 1018405=1009°+324 and the ap- 
proximation 


sts 


24 
It was here necessary that the fraction to replace — should be 


: : bY (Oy Ve 
greater but approximately equal to it, and δ satisfied the conditions, 


while the subsequent work did not require any change in it. 


Ixxxvlil INTRODUCTION. 


(4) »/40692842,. 
The usual process gave 4069284. = 2017? + 995, ; it followed 
that 
36.995 +1] ποσπσςτπτ 
a ao py > /4069284,2,, 
and 20174 was an obvious value to take as an approximation 
somewhat greater than the left side of the inequality. 


(5) /349450. 

In the case of this and the two following roots an approximation 
had to be obtained which was /ess, instead of greater, than the true 
value. Thus Archimedes had to use the second part of the formula 


b 


b 
at 5,70 tb>aty iy: 


In the particular case of ./349450 the integral part of the root is 
591, and the remainder is 169. This gave the result 


16 169 
B91 +a 501 2.59141’ 


and since 160-- 18", while 2.591+1=7.13°, it resulted without 
further calculation that 


> /349450 > 5914 


/349450 > 6911. 

Why then did Archimedes take, instead of this approximation, 
another which was not so close, viz. 59142 The answer which the 
subsequent working and the other approximations in the first part of 
the proof suggest is that he preferred, for convenience of calculation, 

1 

to use for his approximations fractions of the form = δὴ only. But he 
could not have failed to see that to take the nearest fraction of this 
aes iL ae : : 
form, 3? instead of z might conceivably affect his final result and 
make it less near the truth than it need be. As a matter of fact, 
as Hultsch shows, it does not affect the result to take 5914 and to 
work onwards from that figure. Hence we must suppose that 
Archimedes had satisfied himself, by taking 5914 and proceeding on 
that basis for some distance, that he would not be introducing any 
appreciable error in taking the more convenient though less accurate 
approximation 5911. 


ARITHMETIC IN ARCHIMEDES. Ixxxix 


(6) ./137394333, 


In this case the integral portion of the root is 1172, and the 
remainder 35923. Thus, if & denote the root, 


37 Se 
79 64 
AMORA. 
359 
τς τ τος τ lore. 
>117 Ἐς Τ|75 11’ a fortiorr 
Now 2.1172+1=2345; the fraction accordingly becomes 9345? 
and : (= ima) satisfies the necessary conditions, viz. that it must 


be approximately equal to, but not greater than, the given fraction. 
Here again Archimedes would have taken 11721 as the approximate 
value but that, for the same reason as in the last case, 11721 was 
more convenient. 


(7) ./547213255 


The integral portion of the root is here 2339, and the remainder 
1211,4,, so that, if & is the exact root, 


12113, 
2. 233941 
> 23391, a fortiori. 


A few words may be added concerning Archimedes’ ultimate 
reduction of the inequalities 


6671 284 
»π:: 9- 


* 46734 2017 


to the simpler result 3 : >1r>3 τ ‘ 
1 6674 

As a matter of fact ἘΝ 46725" 

only necessary to make the small change of diminishing the de- 


R> 2339 + 


so that in the first fraction it was 


nominator by | in order to obtain the simple ae 


2844 1137, 
2017} ~ 8069° 
Hultsch ingeniously suggests the method of trying the effect of 
increasing the denominator of the latter fraction by 1. This 


As regards the lower limit for 7, we see that and 


ΧΟ INTRODUCTION. 


τς a ; and, if we divide 2690 by 379, the quotient 
is between 7 and 8, so that 
Le ἢ 
7 ~ 2690 ~ 8° 
Now it is a known proposition (proved in Pappus vil. p. 689) 


a@ at+e 
that, if 5 then ho bad 


Similarly it may be proved that 


produces 


ate 6 
b+d a’ 
It follows in the above case that 
919... (STO il 
2690 2690. 8 8” 
which exactly gives = > ! 


10; 379 1. 
and -- 7] is very much nearer to 5,5. 3690 than Ὁ 


Note on alternative hypotheses with regard to the 


approximations to /3. 


For a description and examination of all the various theories put 
forward, up to the year 1882, for the purpose of explaining Archimedes’ 


approximations to 1/3 the reader is referred to the exhaustive paper by 
Dr Siegmund Giinther, entitled Die guadratischen Irrationalitdten der Alten 
und deren Entwickelungsmethoden (Leipzig, 1882). The same author gives 
further references in his Abriss der Geschichte der Mathematik und der Natur- 
wissenschaften im Altertum forming an Appendix to Vol. v. Pt. 1 of Iwan von 
Miiller’s Handbuch der klassischen Altertums-wissenschaft (Miinchen, 1894). 
Giinther groups the different hypotheses under three general heads : 


(1) those which amount to a more or less disguised use of the 
method of continued fractions and under which are included the solutions 
of De Lagny, Mollweide, Hauber, Buzengeiger, Zeuthen, P. Tannery (first 
solution), Heilermann ; 


(2) those which give the approximations in the form of a series 


of fractions such as ὦ + : -- : a 1 +...; under this class come the 


Gt 919s Sass 
solutions of Radicke, v. Pessl, Rodet (with reference to the Qulvasiitras), 


Tannery (second solution) ; 


ARITHMETIC IN ARCHIMEDES. XCl 


(3) those which locate the incommensurable surd between a greater 
and lesser limit and then proceed to draw the limits closer and closer. 
This class includes the solutions of Oppermann, Alexejeff, Schénborn, 
Hunrath, though the first two are also connected by Giinther with the 
method of continued fractions. 

Of the methods so distinguished by Giinther only those need be here 
referred to which can, more or less, claim to rest on a historical basis 
in the sense of representing applications or extensions of principles laid 
down in the works of Greek mathematicians other than Archimedes which 
have come down to us. Most of these quasi-historical solutions connect 
themselves with the system of side- and diagonal-numbers (πλευρικοὶ and 
διαμετρικοὶ ἀριθμοί) explained by Theon of Smyrna (c. 130 A.D.) in a work 
which was intended to give so much of the principles of mathematics as 
was necessary for the study of the works of Plato. 

The stde- and diagonal-numbers are formed as follows. We start with 
two units, and (a) from the sum of them, (Ὁ) from the sum of twice 
the first unit and once the second, we form two new numbers ; thus 


1.141=2, 2.141=3. 


Of these numbers the first is a side- and the second a diagonal-number 
respectively, or (as we may say) 
ας ΞΞ2, d,.=3. 
In the same way as these numbers were formed from a,=1, d,=1, suc- 
cessive pairs of numbers are formed from a,, d,, and so on, in accordance 
with the formula 
An+1=AIntn, Ay 41 = 20, +n, 
whence we have 
d,=1.243=5, d,5=2.24+3=7, 
=1.5+7=12, dy=2.54+7=17, 
and so on. 
Theon states, with reference to these numbers, the general proposition 
which we should express by the equation 
dy? =D? +1. 
The proof (no doubt omitted because it was well-known) is simple. For 
we have 
Oy? — Qty? = (2d, 1 + Ay 1)? — 2 (Gin - +A)? 

Ξε2α,.. i a An" 

Sle (d,-?= 2a,_1”) 

= + (d,_.? — 2a,-,”), and so on, 


while d,?—2a,?= --Ἰ ; whence the proposition is established. 

Cantor has pointed out that any one familiar with the truth of this 
proposition could not have failed to observe that, as the numbers were 
successively formed, the value of d,?/a,? would approach more and more 
nearly to 2, and consequently the successive fractions d,/a, would give 


xcll INTRODUCTION. 


nearer and nearer approximations to the value of 4/2, or in other words that 
ego ert i Medi 
ΤᾺ 2? 5? 12° 29’ eeeece 
are successive approximations to »/2. It is to be observed that the third 
of these approximations, . is the Pythagorean approximation which 


appears to be hinted at by Plato, while the above scheme of Theon, 
amounting to a method of finding all the solutions in positive integers of 
the indeterminate equation 
272 — y= +1, 

and given in a work designedly introductory to the study of Plato, 
distinctly suggests, as Tannery has pointed out, the probability that even 
in Plato’s lifetime the systematic investigation of the said equation had 
already begun in the Academy. In this connexion Proclus’ commentary 
on Eucl. 1. 47 is interesting. It is there explained that in isosceles 
right-angled triangles “it is not possible to find numbers corresponding to 
the sides; for there is no square number which is double of a square 
except in the sense of approaimately double, e.g. 7? is double of 5? less 1.” 
When it is remembered that Theon’s process has for its object the finding 
of any number of squares differing only by unity from double the squares 
of another series of numbers respectively, and that the sides of the two 
sets of squares are called diagonal- and side-numbers respectively, the 
conclusion becomes almost irresistible that Plato had such a system in 
mind when he spoke of ῥητὴ διάμετρος (rational diagonal) as compared 
with ἄρρητος διάμετρος (irrational diagonal) τῆς πεμπάδος (cf. p. xxviii above). 

One supposition then is that, following a similar line to that by which 
successive approximations to »/2 could be obtained from the successive 
solutions, in rational numbers, of the indeterminate equations 247 —y?= +1, 
Archimedes set himself the task of finding all the solutions, in rational 
numbers, of the two indeterminate equations bearing a similar relation 
to 3, viz. 

v*—37?=1, 
et —3y?= -- 2. 

Zeuthen appears to have been the first to connect, eo nomine, the ancient 
approximations to /3 with the solution of these equations, which are also 
made by Tannery the basis of his first method. But, in substance, the 
same method had been used as early as 1723 by De Lagny, whose 
hypothesis will be, for purposes of comparison, described after Tannery’s 
which it so exactly anticipated. 


Zeuthen’s solution. 

After recalling the fact that, even before Euclid’s time, the solution 
of the indeterminate equation #?+72=2 by means of the substitutions 
m? —n* m+n 


L=MN, eae 5 


ARITHMETIC IN ARCHIMEDES, xclli 


was well known, Zeuthen concludes that there could have been no 
difficulty in deducing from Eucl. τι. 5 the identity 


m?—3n?\?__ (m?+3n*\* 
3 (nn)?+( 3 ) - 5) ) ) 
from which, by multiplying up, it was easy to obtain the formula 
3 (Qmn)? + (m? — 8n?)? = (m? + 3x7). 


If therefore one solution m?—3n?=1 was known, a second could at once 
be found by putting 


L=m?+3n7, y=2mn. 
Now obviously the equation 
m —3n?=1 


is satisfied by the values m=2, n=1; hence the next solution of the 
equation 
xv? — 3y?=1 


is =274+3.1=7, yy=2.2.1=4; 
and, proceeding in like manner, we have any number of solutions as 
Bye=P4+3.4=97, youd.7.4=56, 


24=9724+3.562=18817, yg=2.97.56=10864, 
and so on. 


Next, addressing himself to the other equation 


ue? — 3y?= — 2, 
Zeuthen uses the identity 


(m+3n)?—3 (m+n)?= — 2 (m?—3n?). 
Thus, if we know one solution of the equation m?—3n?=1, we can proceed 


to substitute 
L=M+3n, Y=M+N. 


Suppose m=2, n=1, as before ; we then have 

a,=5, Yy=3. 
If we put #,=2,+3y,=14, y,=27,+4,=8, we obtain 
a es 


ee ὃ 4 
(and m=7, n=4 is seen to be a solution of m?—3n?=1). 


Starting again from 7, y., we have 
,=38, Y3= 22, 


and zs 


whence “ἜΞΕ - 


XClv INTRODUCTION. 


(and m=26, n=15 satisfies m?-3n?=1), 
,=284,  y,=164, 


or ὅς εἶθ 
Ye ΧΑ 
Similarly “= ee Ete ΞΘΟ and isos 00: 


Ys 56’ y 153’ 
This method gives all the successive approximations to /3, taking 

account as it does of both the equations 

v—3y=1, 

φῇ — 3y?= — 2. 


Tannery's first solution. 


Tannery asks himself the question how Diophantus would have set 
about solving the two indeterminate equations. He takes the first equation 
in the generalised form 

v?—ay*=1, 
and then, assuming one solution (p, 4) of the equation to be known, he 
supposes 
»)ιτεηιῶ-- , H=r+q- 


Then py — ag? =m" — Impx +p? — ax? —2age—ag?=1, 
whence, since p?—aq?=1, by hypothesis, 
o=2,—P ΞΙ 51 ᾿ 
m — a 
, _ (m7? +a) p+2amq _ 2mp+(m?+a)q 
so that i=" gee Vie 


and p,?—aq,?=1. 
The values of p,, g, so found are rational but not necessarily integral ; 
if integral solutions are wanted, we have only to put 


P= (+ av?) p+ 2aurg, 9, =2pu + (uv? + av") g, 
where (εν, v) is another integral solution of #? -- ay?=1. 
Generally, if (p, g) be a known solution of the equation 
v—ay’=r, 
suppose p,=ap+q, g,=yp + 6g, and “il suffit pour déterminer a, β, y, ὃ de 
connaitre les trois groupes de solutions les plus simples et de résoudre 


deux couples d’équations du premier degré ἃ deux inconnues.” Thus 


(1) for the equation 
v?—3y?=1, 
the first three solutions are 


(p=, q=9), (p=2, g=1), (p=7, q=4), 


2=a 7=2a+fB 
whence 152. and ray Ae 


so that a=2, B=3, y=1, 8=2, 


ARITHMETIC IN ARCHIMEDES. XCV 


and it follows that the fourth solution is given by 
p=2.7+3.4=26, 
g=1.7+2.4=15; 
(2) for the equation x? —37?= --2, 
the first three solutions being (1, 1), (5, 3), (19, 11), we have 
= 19=5a+3 
oat and iy ene , 
whence a=2, B=3, y=1, 5=2, and the next solution is given by 
p=2.19+3.11=71, 


g=1.19+2.11=41, 
and so on. 
Therefore, by using the two indeterminate equations and proceeding as 
shown, all the successive approximations to V3 can be found. 
Of the two methods of dealing with the equations it will be seen that 
Tannery’s has the advantage, as compared with Zeuthen’s, that it can be 
applied to the solution of any equation of the form x?—ay*=r. 


De Lagny’s method. 


The argument is this. If /3 could be exactly expressed by an im- 
proper fraction, that fraction would fall between 1 and 2, and the square of 
its numerator would be three times the square of its denominator. Since 
this is impossible, two numbers have to be sought such that the square of 
the greater differs as little as possible from 3 times the square of the 
smaller, though it may be either greater or less. De Lagny then evolved 
the following successive relations, 


2=3.124+1, 57=3.3?-2 72=3.4?+1, 19%=3.11?-2, 
267=3.157+1, 712=3. 412-2, ete. 


From these relations were derived a series of fractions greater than V3, 


viz. = a: = , etc., and another series of fractions less than V3, viz. 
5 19 7] : : 3 
3°11’ 41’ etc. The law of formation was found in each case to be that, if 


2 was one fraction in the series and x the next, then 


ΤΡ Ἐ50 
q 5) 
This led to the results 
ST F2G τον 1862 1351 fe 
1747 15. 56> 9090. 780 «τὸ ονο, 
δ. 19 71 905 989 369] = 
and 5. τὶ <a < 153 “671 9151" <3 


xevl INTRODUCTION. 


while the law of formation of the successive approximations in each series 
is precisely that obtained by Tannery as the result of treating the two 
indeterminate equations by the Diophantine method. 


Heilermann’s method. 


This method needs to be mentioned because it also depends upon a 
generalisation of the system of séde- and diagonal-numbers given by Theon 
of Smyrna. 

Theon’s rule of formation was 


Sn=Sn-1+Da—1; Dy, =28p-1+ Dn-13 
and Heilermann simply substitutes for 2 in the second relation any 
arbitrary number a, developing the following scheme, 


S,=Sot+ Do; D, = aSy+ Dy; 
S,=+D,, D,=aS,+D,, 
S3=S,+D,, D;=aS,.+D,, 


ΘΙ ΞΡ ϑΞδ ΓΕ 
It follows that 
AS? = aS yy? + 2a8,_; Dn_ + aD y_4*, 
De =O Byatt Oe 
By subtraction, D,2 — aS,2= (1 — a) (Dy_ 12 — αϑ...2) 
(1 —a)? (Dy? — an”), similarly, 


=(1—a)" (D,?- a8,?). 
This corresponds to the most general form of the “ Pellian” equation 
43 — ay? = (const.). 
If now we put D)=S)=1, we have 
D,?2 (. -- αὐ. 11 
ἜΣ ὍΣ 
from which it appears that, where the fraction on the right-hand side 


. ἌΣ : aie 
approaches zero as ” increases, y 15 an approximate value for Va. 
n 


Clearly in the case where a=3, Dy)=2, Sp=1 we have 


Dy2 Τὶ δος, DO aes 


Bi? & 3? Sy) Gee? 6, gale Ib? 
D, Te VD: let or ae seen 


ΠῚ 


S, ἀπ δ. dle 60” 82 v3" 


and so on, 


ARITHMETIC IN ARCHIMEDES. X¢eV1l 
But the method is, as shown by Heilermann, more rapid if it is used to 
find, not γα, but b/a, where ὁ is so chosen as to make 0?a (which takes 


the place of a) somewhat near to unity. Thus suppose a=2, so that 


Va=2 V3, and we then have (putting D)=S)=1) 


52 5 26 26 
S,=2, “Ὁ and V30 5. Ree ge 

102 54452 106 106, 265 
mon ar ean π᾿ and 4/8 os 3 102? * 153° 
208 102.27 106 5404 
Ss=o5 Ys 95.95 +95 85. 58’ 


- 5404 ὅ 1351 
SIG regener or 780 ° 
This is one of the very few instances of success in bringing out the two 
Archimedean approximations in immediate sequence without any foreign 
values intervening. No other methods appear to connect the two values 
in this direct way except those of Hunrath and Hultsch depending on the 
formula 


and 


b τ ὖ 
ats >V/a+b HOS ab Ἐπ 


We now pass to the second class of solutions which develops the 
approximations in the form of the sum of a series of fractions, and under 
this head comes 

Tannery's second method. 

This may be exhibited by means of its application (1) to the case of the 
square root of a large number, e.g. /349450 or 5712+ 23409, the first of 
the kind appearing in Archimedes, (2) to the case of 4/3. 

(1) Using the formula 


ΠΕΣ ΤΟΝ 
2a 


we try the effect of putting for /571?+ 23409 the expression 


23409 
1142 " 


It turns out that this gives correctly the integral part of the root, and we 
now suppose the root to be 


571 + ——— 


1 
571+20+—. 
m 
Squaring and regarding = as negligible, we have 


7124400 + 22840+ ae ar Ὁ 5717+ 23409, 


xXe@Vlil INTRODUCTION. 


whence Lipet 169, 
m 
and ΘΗΝ Ss 
he 1859. 7 
so that /349450 >591 τ 
(2) Bearing in mind that 
b 
2 
V@+beate ἘΠ 
ἈΞ 2 
pee 2 aie 
we have V/3="/1 +2e01+5 τὶ 
2 5 
~ 1.5» or 3° 


: = 1 ς 
Assuming then that /3 = G + = , Squaring and neglecting = , we obtain 


engl 
St sa 
whence m=15, and we get as the second approximation 
Bey re ae 
3. 15’ ES 
We have now 26?-3.15?=1, 


and can proceed to find other approximations by means of Tannery’s first 
method. 


ΘΠ = eal NS 
Or we can also put (+3+ists) =) 


and, neglecting =) we get 
267 ὅδ᾽ 
ΤΙΣ. Abia: 
whence n= —15.52= — 780, and 


Ξ Ἢ 1 1351 
v3.09 (145 +35 - 7607 780)” 


It is however to be observed that this method only connects a with 


and not with the intermediate approximation τ to obtain which 


Tannery implicitly uses a particular case of the formula of Hunrath and 
Hultsch. 


Rodet’s method was apparently invented to explain the approximation 
in the Culvasitras* 
1 1 


VB 1+5 3134 39,4584" 


* See Cantor, Vorleswngen tiber Gesch. d. Math. Ὁ. 600 sq. 


ARITHMETIC IN ARCHIMEDES. ΧΟΙΧ 


but, given the approximation 5 2 , the other two successive approximations 
indicated by the formula can be obtained by the method of squaring just 
described* without such elaborate work as that of Rodet, which, when 
applied to 4/3, only gives the same results as the simpler method. 
Lastly, with reference to the third class of solutions, it may be 
mentioned 
(1) that Oppermann used the formula 
at 2ab 
—>V/ab >— τη} 


which gave successively 


ΙΝ 
but only led to one of the Archimedean approximations, and that by 
combining the last two ratios, thus 

97+168 265 

56497 153? 
(2) that Schénborn came somewhat near to the formula successfully used 
by Hunrath and Hultsch when he proved t te 


ate PNG EIS a+ 


ai 


* Cantor had already pointed this out in his first edition of 1880. 
+ Zeitschrift fiir Math. u. Physik (Hist. litt, Abtheilung) xxviu. (1883), 
p. 169 sq. 


CHAPTER V. 
ON THE PROBLEMS KNOWN AS ΝΕΥΣΕΙΣ. 


THE word νεῦσις, commonly znclinatio in Latin, is difficult to 
translate satisfactorily, but its meaning will be gathered from some 
general remarks by Pappus having reference to the two Books of 
Apollonius entitled νεύσεις (now lost). Pappus says*, “A line is 
said to verge (vevew) towards a point if, being produced, it reach the 
point,” and he gives, among particular cases of the general form of 
the problem, the following. 

“Two lines being given in position, to place between them a 
straight line given in length and verging towards a given point.” 

“Tf there be given in position (1) a semicircle and a straight 
line at right angles to the base, or (2) two semicircles with their 
bases in a straight line, to place between the two lines a straight 
line given in length and verging towards a corner (ywviav) of a 
semicircle.” 

Thus a straight line has to be laid across two lines or curves so 
that it passes through a given point and the intercept on it between 
the lines or curves is equal to a given lengthf. 


81. The following allusions to particular vevoes are found in 
Archimedes. The proofs of Props. 5, 6, 7 of the book On Spirals 
use respectively three particular cases of the general theorem that, 


* Pappus (ed. Hultsch) vit. p. 670. 

+ In the. German translation of Zeuthen’s work, Die Lehre von den 
Kegélschnitten im Altertum, νεῦσις is translated by “ Kinschiebung,”’ or as we 
might say “insertion,” but this fails to express the condition that the required 
line must pass through a given point, just as inclinatio (and for that matter the 
Greek term itself) fails to express the other requirement that the intercept on 
the line must be of given length. 


ON THE PROBLEMS KNOWN AS ΝΕΥΣΕΙΣ, cl 


if A be any point on a circle and BC any diameter, ἐξ is possible to 
draw through A a straight line, meeting the circle again in P and 
BC produced in R, such that the intercept PR is equal to any given 


length. In each particular case the fact is merely stated as true 
without any explanation or proof, and 


(1) Prop. 5 assumes the case where the tangent at A is parallel 
to BC, 


(2) Prop. 6 the case where the points A, P in the figure are 
interchanged, 


(3) Prop. 7 the case where A, P are in the relative positions 
shown in the figure. 

Again, (4) Props. 8 and 9 each assume (as before, without proof, 
and without giving any solution of the 
implied problem) that, if AZ, BC be two 
chords of a circle intersecting at right 
angles in a point D such that BD> DC, 
then it is possible to draw through A 
another line ARP, meeting BC in R and 
the circle again in P, such that PR = DE. 

Lastly, with the assumptions in Props. 
ὅ, 6, 7 should be compared Prop. 8 of the 
Liber Assumptorum, which may well be 
due to Archimedes, whatever may be said of the composition of the 
whole book. This proposition proves that, if in the first figure 
APR is so drawn that PR is equal to the radius OP, then the are 
AB is three times the are PC. In other words, if an arc AB of a 
circle be taken subtending any angle at the centre QO, an arc equal 
to one-third of the given arc can be found, ἐ.6. the given angle can be 
trisected, uf only APR can be drawn through A in such a manner 


cll INTRODUCTION. 


that the intercept PR between the circle and BO produced is equal to 
the radius of the circle. Thus the trisection of an angle is reduced to 
a νεῦσις exactly similar to those assumed as possible in Props. 6, 7 
of the book On Spirals. 

The νεύσεις so referred to by Archimedes are not, in general, 
capable of solution by means of the straight line and circle alone, 
as may be easily shown. Suppose in the first figure that x 
represents the unknown length OF, where O is the middle point 
of BOC, and that & is the given length to which PF is to be equal ; 
also let OD=a, AD=b, BC =2c. Then, whether BC be a diameter 
or (more generally) any chord of the circle, we have 


Ah te — Bi, 
and therefore kJ? + (α-- α}"- οὗ -- ὁ, 


The resulting equation, after rationalisation, is an equation of the 
fourth degree in x; or, if we denote the length of AA by y, we have, 
for the determination of x and y, the two equations 


y? = (α -- a)? + δ᾽ 

ky = α -- οὗ } 
In other words, if we have a rectangular system of coordinate 
axes, the values of x and y satisfying the conditions of the problem 
can be determined as the coordinates of the points of intersection of 
a certain rectangular hyperbola and a certain parabola. 

In one particular case, that namely in which D coincides with O 
the middle point of BC, or in which A is one extremity of the 
diameter bisecting BC at right angles, a=0, and the equations 
reduce to the single equation 


y’ —ky=b' -- οὗ, 
which is a quadratic and can be geometrically solved by the 


traditional method of application of areas; for, if οὐ be substituted 
for y—k, so that «= AP, the equation becomes 


αν (ἢ - τι) τε δ᾽ +e’, 


and we have simply “to apply to a straight line of length & a 
rectangle exceeding by a square figure and equal to a given 
area (b+ c?).” 

The other νεῦσις referred to in Props. 8 and 9 can be solved in 
the more general form where ὦ, the given length to which PA 
is to be equal, has any value within a certain maximum and is not 


ON THE PROBLEMS KNOWN AS ΝΕΎΣΕΙΣ. ΟἿ 


necessarily equal to DZ, in exactly the same manner ; and the two 
equations corresponding to (a) will be for the second figure 


x 


y? -- (α -- αν)" +b? 
ly =e Ὡς j AERA ERE λας (B). 

Here, again, the problem can be solved by the ordinary method 
of application of areas in the particular case where AJL is the 
diameter bisecting BC at right angles; and it is interesting to note 
that this particular case appears to be assumed in a fragment 
of Hippocrates’ Quadrature of lunes preserved in a quotation 
by Simplicius* from Eudemus’ History of Geometry, while Hippo- 
crates flourished probably as early as 450 B.c. 

Accordingly we find that Pappus distinguishes different classes 
of νεύσεις corresponding to his classification of geometrical problems 
in general. According to him, the Greeks distinguished three kinds 
of problems, some being plane, others solid, and others linear. He 
proceeds thus7: ‘‘ Those which can be solved by means of a straight 
line and a circumference of a circle may properly be called plane 
(ἐπίπεδα) ; for the lines by means of which such problems are 
solved have their origin in a plane. Those however which are 
solved by using for their discovery (εὕρεσιν) one or more of the 
sections of the cone have been called solid (στερεά); for the 
construction requires the use of surfaces of solid figures, namely, 
those of cones. There remains a third kind of problem, that — 
which is called linear (γραμμικόν) ; for other lines [curves] besides ἡ 
those mentioned are assumed for the construction whose origin 
is more complicated and less natural, as they are generated from 
more irregular surfaces and intricate movements.” Among other 
instances of the linear class of curves Pappus mentions spirals, the 
curves known as quadratrices, conchoids and cissoids. He adds 
that “it seems to be a grave error which geometers fall into 
whenever any one discovers the solution of a plane problem by 
means of conics or linear curves, or generally solves it by means of 
a foreign kind, as is the case, for example, (1) with the problem in 
the fifth Book of the Conics of Apollonius relating to the parabola}, 

* Simplicius, Comment. in Aristot. Phys. pp. 61—68 (ed. Diels). The whole 
quotation is reproduced by Bretschneider, Die Geometrie und die Geometer vor 
Euklides, pp. 109—121. As regards the assumed construction see particularly 
p- 64 and p. xxiv of Diels’ edition; cf. Bretschneider, pp. 114,115, and Zeuthen, 
Die Lehre von den Kegelschnitten im Altertum, pp. 269, 270. 


+ Pappus rv. pp. 270—272. 
1 Cf. Apollonius of Perga, pp. exxvili. exxix. 


οἷν INTRODUCTION. 


and (2) when Archimedes assumes in his work on the spiral a 
νεῦσις Of a solid character with reference to a circle; for it is 
possible without calling in the aid of anything solid to find the 
[proof of the] theorem given by the latter [Archimedes], that is, to 
prove that the circumference of the circle arrived at in the first 
revolution is equal to the straight line drawn at right angles to the 
initial line to meet the tangent to the spiral.” 

The “solid vetous” referred to in this passage is that assumed to 
be possible in Props. 8 and 9 of the book On Spirals, and is mentioned 
again by Pappus in another place where he shows how to solve the 
problem by means of conics*. This solution will be given later, but, 
when Pappus objects to the procedure of Archimedes as unorthodox, 
the objection appears strained if we consider what precisely it is that 
Archimedes assumes. It is not the actual solution which is assumed, 
but only its possibility; and its possibility can be perceived without 
any use of conics. For in the particular case it is only necessary, 
as a condition of possibility, that DZ# in the second figure above 
should not be the maximum length which the intercept PR could 
have as APR revolves about A from the position ADF in the 
direction of the centre of the circle; and that DE is not the 
maximum length which PA can have is almost self-evident. In 
fact, if P, instead of moving along the circle, moved along the 
straight line through £ parallel to LC, and if ARP moved from the 
position ADF in the direction of the centre, the length of P# would 
continually increase, and ὦ fortiori, so long as P is on the are of the 
circle cut off by the parallel through # to BC, PR must be greater 
in length than D#; and on the other hand, as A4#P moves further 
in the direction of 5, it must sometime intercept a length PR 
equal to DE before P reaches 4, when P# vanishes. Since, then, 
Archimedes’ method merely depends upon the theoretical possibility 
of a solution of the νεῦσις, and this possibility could be inferred 
from quite elementary considerations, he had no occasion to use 
conic sections for the purpose immediately in view, and he cannot 
fairly be said to have solved a plane problem by the use of conics. 

At the same time we may safely assume that Archimedes 
was in possession of a solution of the νεῦσις referred to. But there 
is no evidence to show how he solved it, whether by means of conics, 
or otherwise. That he would have been able to effect the solution, 


* Pappus Iv. p. 298 sq. 


ON THE PROBLEMS KNOWN AS ΝΕΥΣΕΙΣ, CV 


as Pappus does, by the use of conics cannot be doubted. A precedent 
for the introduction of conics where a “solid problem” had to be 
solved was at hand in the determination of two mean proportionals 
between two unequal straight lines by Menaechmus, the inventor οὗ 
the conic sections, who used for the purpose the intersections of a 
parabola and a rectangular hyperbola. The solution of the cubic 
equation on which the proposition On the Sphere and Cylinder τι. 4 
depends is also effected by means of the intersections of a parabola 
with a rectangular hyperbola in the fragment given by Eutocius 
and by him assumed to be the work of Archimedes himself*. 

Whenever a problem did not admit of solution by means of the 
straight line and circle, its solution, where possible, by means of 
conics was of the greatest theoretical importance. First, the 
possibility of such a solution enabled the problem to be classified 
as a “solid problem”; hence the importance attached by Pappus 
to solution by means of conics. But, secondly, the method had 
other great advantages, particularly in view of the requirement that 
the solution of a problem should be accompanied by a διορισμός 
giving the criterion for the possibility of a real solution. Often too 
the διορισμός involved (as frequently in Apollonius) the determination 
of the number of solutions as well as the limits for their possibility. 
Thus, in any case where the solution of a problem depended on the 
intersections of two conics, the theory of conics afforded an effective 
means of investigating διορισμοί. 


§ 2. But though the solution of “solid problems” by means of 
conics had such advantages, it was not the only method open to 
Archimedes. An alternative would be the use of some mechanical 
construction such as was often used by the Greek geometers and is 
recognised by Pappus himself as a legitimate substitute for conics, 
which are not easy to draw in a planet. Thus in Apollonius’ 
solution of the problem of the two mean proportionals as given by 
Eutocius a ruler is supposed to be moved about a point until the 
points at which the ruler crosses two given straight lines at right 
angles are equidistant from a certain other fixed point; and the 
same construction is also given under Heron’s name. Another 
version of Apollonius’ solution is that given by Ioannes Philoponus, 
which assumes that, given a circle with diameter OC and two 


* See note to On the Sphere and Cylinder, τι. 4. 
+ Pappus ut. p. 54, 


ronal INTRODUCTION. 


straight lines OD, OF through O and at right angles to one 
another, a line can be drawn through C, meeting the circle again 
in #/ and the two lines in D, £ respectively, such that the in- 
tercepts CD, FE are equal. This solution was no doubt discovered 
by means of the intersection of the circle with a rectangular hyper- 
bola drawn with OD, OF as asymptotes and passing through C ; 
and this supposition accords with Pappus’ statement that Apollonius 
solved the problem by means of the sections of the cone*. The 
equivalent mechanical construction is given by Eutocius as that 
of Philo Byzantinus, who turns a ruler about C until CD, FEF are 
equal 7. 

Now clearly a similar method could be used for the purpose of 
effecting a νεῦσις. We have only to suppose a ruler (or any object 
with a straight edge) with two marks made on it at a distance 
equal to the given length which the problem requires to be 
intercepted between two curves by a line passing through the 
fixed point; then, if the ruler be so moved that it always passes 
through the fixed point, while one of the marked points on it follows 
the course of one of the curves, it is only necessary to move the 
ruler until the second marked point falls on the other curve. Some 
such operation as this may have led Nicomedes to the discovery of 
his curve, the conchoid, which he introduced (according to Pappus) 
into his doubling of the cube, and by which he also trisected an 
angle (according to the same authority). From the fact that 
Nicomedes is said to have spoken disrespectfully of Eratosthenes’ 
mechanical solution of the duplication problem, and therefore must 
have lived later than Eratosthenes, it is concluded that his date 
must have been subsequent to 200 B.c., while on the other hand 
he must have written earlier than 70 B.c., since Geminus knew the 
name of the curve about that date; Tannery places him between 
Archimedes and Apolloniust. While therefore there appears to 
be no evidence of the use, before the time of Nicomedes, of such 
a mechanical method of solving a νεῦσις, the interval between 
Archimedes and the discovery of the conchoid can hardly have 
been very long. As a matter of fact, the conchoid of Nicomedes 
can be used to solve not only all the νεύσεις mentioned in Archimedes 
but any case of such a problem where one of the curves is a straight 


* Pappus 111. p. ὅθ. 
+ For fuller details see Apollonius of Perga, pp. ¢XXV—CXxvil. 
1 Bulletin des Sciences Mathématiques, 2° série vit. p. 284, 


ON THE PROBLEMS KNOWN AS ΝΕΎΣΕΙΣ. evil 


line. Both Pappus and Eutocius attribute to Nicomedes the inven- 
tion of a machine for drawing his conchoid. AP is supposed to be 


a ruler with a slot in it parallel to its length, /# a second ruler at 
right angles to the first with a fixed peg in it, C. This peg moves 
in a slot made in a third ruler parallel to its length, while this 
ruler has a fixed peg on it, D, in a straight line with the slot in 
which (Οὐ moves ; and the peg D can move along the slot in Ab. Τί 
then the ruler PD moves so that the peg D describes the length of 
the slot in AB on each side of /, the extremity of the ruler, P, 
describes the curve which is called a conchoid. Nicomedes called 
the straight line AB the ruler (κανών), the fixed point C the pole 
(πόλος), and the length PD the distance (διάστημα) ; and the 
fundamental property of the curve, which in polar coordinates 
would now be denoted by the equation r=a+bsec θ, is that, if 
any radius vector be drawn from C to the curve, as CP, the length 
intercepted on the radius vector between the curve and the straight 
line AB is constant. Thus any νεῦσις in which one of the two 
given lines is a straight line can be solved by means of the 
intersection of the other line with a certain conchoid whose pole 
is the fixed point to which the required straight line must verge 
(vevew). In practice Pappus tells us that the conchoid was not 
always actually drawn, but that “some,” for greater convenience, 
moved the ruler about the fixed point until by trial the intercept 
was made equal to the given length*. 


§ 3. The following is the way in which Pappus applies 
conic sections to the solution of the «νεῦσις referred to in Props. ὃ, 9 
of the book On Spirals. He begins with two lemmas. 


* Pappus Iv. p. 246. 


ΟΥ̓ΠῚ INTRODUCTION. 


(1) If from a given point A any straight line be drawn meeting 
a straight line BC given in position in #, and if RY be drawn 
perpendicular to BC and bearing a given ratio to AR, the locus of 
Q is a hyperbola. 


For draw AD perpendicular to BC, and on AD produced take A’ 
such that 
Qk: RA=A'D: DA =(the given ratio). 
Measure DA” along DA equal to DA’. 
Then, if QV be perpendicular to AW, 
(AR? — AD?) ; (QR? — A'D*) = (const.), 
or QN* : A'N . A” N = (const.) 


(2) If BC be given in length, and if RQ, a straight line drawn 
at right angles to BC from any point 1 on it, be such that 
BR. RC=k. RQ, 
where & is a straight line of given length, then the locus of Q isa 
parabola. 


Let O be the middle point of BC, and let OX be drawn at right 
angles to it and of such length that 


OC* =k. KO. 
Draw QN’ perpendicular to OX. 
Then ON? = OR? ΞΟ BE ne 


=k.(KO— RQ), by hypothesis, 
= Kea piel 


ON THE PROBLEMS KNOWN AS ΝΕΥΣΕΙ͂Σ. ΟΙΧ 


In the particular case referred to by Archimedes (with the slight 
generalisation that the given length & to which P£ is to be equal is 
not necessarily equal to D#) we have 


(1) the given ratio RQ: AF is unity, or RY=AR, whence A” 
coincides with A, and, by the first lemma, 
ON? = AN ΑΝ, 
so that Q lies on a rectangular hyperbola. 


(2) BR. RC=AR. RP=k. AR=k. RQ, and, by the second 
lemma, Q lies on a certain parabola. 

If now we take O as origin, OC as axis of # and OK as axis of y, 
and if we put OD=a, AD=b, BC = 2c, the hyperbola and parabola 
determining the position of ᾧ are respectively denoted by the 
equations 

(α -- α)" -Ξ- γ΄ — δ', 

οὗ --αὐ -- ky, 
which correspond exactly to the equations (8) above obtained by 
purely algebraical methods. 

Pappus says nothing of the διορισμός which is necessary to the 
complete solution of the generalised problem, the διορισμός namely 
which determines the maximwm value of & for which the solution is 
possible, This maximum value would of course correspond to the 
case in which the rectangular hyperbola and the parabola touch one 
another. Zeuthen has shown* that the corresponding value of & can 
be determined by means of the intersection of two other hyperbolas or 
of a hyperbola and a parabola, and there is no doubt that Apollonius, 
with his knowledge of conics, and in accordance with his avowed 
object in giving the properties useful and necessary for διορισμοί, 
would have been able to work out this particular διορισμός by means 
of conies; but there is no evidence to show that Archimedes investi- 
gated it by the aid of conics, or indeed at all, it being clear, as shown 
above, that it was not necessary for his immediate purpose. 

This chapter may fitly conclude with a description of (1) some 
important applications of νεύσεις given by Pappus, and (2) certain 
particular cases of the same class of problems which are plane, that 
is, can be solved by the aid of the straight line and circle only, and 
which were (according to Pappus) shown by the Greek geometers to 
be of that character. 


* Zeuthen, Die Lehre von den Kegelschnitten im Altertum, pp. 273—5. 


cx INTRODUCTION. 


8.4, One of the two important applications of ‘solid’ νεύσεις was 
discovered by Nicomedes, the inventor of the conchoid, who intro- 
duced that curve for solving a νεῦσις to which he reduced the problem 
of doubling the cube* or (what amounts to the same thing) the finding 
of two mean proportionals between two given unequal straight lines. 

Let the given unequal straight lines be placed at right angles as 
CL, LA. Complete the parallelogram 4 BCL, and bisect AB at D, 
and BC at H. Join LD and produce it to meet CB produced in Z. 
From £ draw EF at right angles to BC, and take a point / on £F 
such that CF is equal to AD. Join HF, and through C draw CG& 
parallel to HF. If we produce BC to λ΄, the straight lines CG, CK 


form an angle, and we now draw from the given point / a straight 
line FGK, meeting CG, CK in G, K respectively, such that the 
intercept G'K is equal to AD or FC. (This is the νεῦσις to which 
the problem is reduced, and it can be solved by means of a conchoid 
with F as pole.) 
Join KL and produce it to meet LA produced in 77. 
Then shall C.K, AM be the required mean proportionals between 
CL, LA, or 
Cl: CK =CK : AM=AM AE: 
We have, by Eucl. 11. 6, 
BK ΚΟ + CE*= ER’. 
If we add EF? to each side, 
BK . KC +CF?=FR’. 


Now, by parallels, 
MA:AB=ML: LK 


= BC CK; 
* Pappus rv. p. 242 sq. and m1. p. 58 sq. ; Eutocius on Archimedes, On the 
Sphere and Cylinder, τι. 1 (Vol. 111. p. 114 sq.) 


ON THE PROBLEMS KNOWN AS ΝΕΥΣΕΙΣ. Cx1 


and, since AB=2AD, and BC =+HC, 
MA AD=HO: CK 


= FG : GK, by parallels, 
whence, componendo, 
MD:AD=FEK : GK. 
But GK = AD; therefore VD = ΚΑ, and MD? = ΚΞ". 


Again, MD*?=BM .MA+ AD’, 
and FK°=BK . KC + CF’, from above, 
while MD?* — FK*, and AD*=CF’,; 
therefore BM .MA=BK . KC. 
Hence CK: MA=BM: BK 
=MA:AL 


= OMEN OG 
that is, DOVCKA~CK MA MA: AL. 


§ 5. The second important problem which can be reduced to 
a ‘solid’ νεῦσις is the trisection of any angle. One method of 
reducing it to ἃ νεῦσις has been mentioned above as following from 
Prop. 8 of the Liber Asswmptorum, This method is not mentioned 
by Pappus, who describes (tv. p. 272 sq.) another way of effecting 
the reduction, introducing it with the words, ‘The earlier 
geometers, when they sought to solve the aforesaid problem about 
the [trisection of the] angle, a problem by nature ‘solid,’ by 
‘plane’ methods, were unable to discover the solution; for they 
were not yet accustomed to the use of the sections of the cone, 
and were for that reason at a loss. Later, however, they trisected 
an angle by means of conics, having used for the discovery of it 
the following νεῦσις." 

The νεῦσις is thus enunciated: Given a rectangle ABCD, let it 
be required to draw through A a straight line AVR, meeting CD in 
Q and BC produced in R, such that the intercept QA is equal to a 
given length, & suppose. 

Suppose the problem solved, QR being equal to k. Draw DP 
parallel to Q# and FP parallel to CD, meeting in P. Then, in the 
parallelogram DR, DP = QR =k. 

Hence P lies on a circle with centre D and radius k. 

Again, by Eucl. 1. 43 relating to the complements of the 
parallelograms about the diagonal of the complete parallelogram, 

BC ..CD=BEK. QD 
ΞΡ ΡΒ: 


} , by parallels, 


CXil INTRODUCTION. 


and, since BC. CD is given, it follows that P lies on a rectangular 
hyperbola with BR, BA as asymptotes and passing through D. 


B Cc R 


Therefore, to effect the construction, we have only to draw this 
rectangular hyperbola and the circle with centre D and radius equal 
to k. The intersection of the two curves gives the point P, and & 
is determined by drawing PR parallel to DC. Thus AQF is found. 

[Though Pappus makes ABCD a rectangle, the construction 
applies equally if ABCD is any parallelogram. | 

Now suppose 4 &C to be any acute angle which it is required to 
trisect. Let AC be perpendicular to BC. Complete the parallelo- 
gram ADBC, and produce DA. 

Suppose the problem solved, and let the angle CBZ be one-third 
of the angle ASC. Let BH meet AC in αὶ and DA produced in F. 
Bisect HF in H, and join AH. 

Then, since the angle ABZ is equal to twice the angle HBC and, 
by parallels, the angles HBC, HFA are equal, 


LABE=22£AFH= 1 AHB, 


Therefore Ais A= A 
and ELF=2HF 
= 24}. 


B Cc 


Hence, in order to trisect the angle ABC, we have only to solve 
the following νεῦσις : Given the rectangle ADBC whose diagonal 


ON THE PROBLEMS KNOWN AS ΝΕΥΣΕΙΣ. cxill 


is AB, to draw through B a straight line BEF, meeting AC in E and 
DA produced in F, such that EF may be equal to twice AB; and this 
νεῦσις is solved in the manner just shown. 

These methods of doubling the cube and trisecting any acute 
angle are seen to depend upon the application of one and the same 
νεῦσις, Which may be stated in its most general form thus. Given 
any two straight lines forming an angle and any fixed point 
which is not on either line, it is required to draw through the 
fixed point a straight line such that the portion of it intercepted 
between the fixed lines is equal to a given length. If AEH, AC be 


the fixed lines and & the fixed point, let the parallelogram ACLD 
be completed, and suppose that LYRA, meeting CA in Q and AZ in 
f, satisfies the conditions of the problem, so that QR is equal to 
the given length. If then the parallelogram CQRP is completed, 
we may regard P as an auxiliary point to be determined in order 
that the problem may be solved ; and we have seen that P can be 
found as one of the points of intersection of (1) a circle with centre 
C and radius equal to &, the given length, and (2) the hyperbola 
which passes through C’ and has DH, DB for its asymptotes. 

It remains only to consider some particular cases of the problem 
which do not require conics for their solution, but are ‘plane’ 
problems requiring only the use of the straight line and circle. 


§ 6. We know from Pappus that Apollonius occupied him- 
self, in his two Books of vetoes, with problems of that type 
which were capable of solution by ‘plane’ methods. As a matter 
of fact, the above νεῦσις reduces to a ‘plane’ problem in the 
particular case where £& lies on one of the bisectors of the angle 
between the two given straight lines, or (in other words) where the 
parallelogram ACBD is a rhombus or a square. Accordingly we 
find Pappus enunciating, as one of the ‘plane’ cases which had 


ἘΠῚ A. h 


ΟΧΙΥ INTRODUCTION. 


been singled out for proof on account of their greater utility for 
many purposes, the following*: Given a rhombus with one side 
produced, to fit into the exterior angle a straight line given in 
length and verging to the opposite angle ; and he gives later on, in 
his lemmas to Apollonius’ work, a theorem bearing on the problem 
with regard to the rhombus, and (after a preliminary lemma) 
a solution of the νεῦσις with reference to a square. 

The question therefore arises, how did the Greek geometers 
discover these and other particular cases, where a problem which 
is in general ‘solid,’ and therefore requires the use of conics (or a 
mechanical equivalent), becomes ‘plane’? Zeuthen is of opinion that 
they were probably discovered as the result of a study of the general 
solution by means of conicst. I do not feel convinced of this, for 
the following reasons. 

(1) The authenticated instances appear to be very rare in 
which we should be justified in assuming that the Greeks used 
the properties of conics, in the same way as we should combine 
and transform two Cartesian equations of the second degree, for 
the purpose of proving that the intersections of two conics also 
lie on certain circles or straight lines. It is true that we may 
reasonably infer that Apollonius discovered by a method of this sort 
his solution of the problem of doubling the cube where, in place 
of the parabola and rectangular hyperbola used by Menaechmus, 
he employs the same hyperbola along with the cicle which passes 
through the points common to the hyperbola and parabola{; but 
in the only propositions contained in his conics which offer an 
opportunity for making a similar reduction§, Apollonius does not 
make it, and is blamed by Pappus for not doing so. In the pro- 
positions referred to the feet of the normals to a parabola drawn 
from a given point are determined as the intersections of the 
parabola with a certain rectangular hyperbola, and Pappus objects 


* Pappus vil. p. 670. 

+ ‘Mit dieser selben Aufgabe ist naimlich ein wichtiges Beispiel dafiir 
verknitipft, dass man bemiiht war solche Falle zu entdecken, in denen Aufgaben, 
zu deren Lésung im allgemeinen Kegelschnitte erforderlich sind, sich mittels 
Zirkel und Lineal lésen lassen. Da nun das Studium der allgemeinen Lésung 
durch Kegelschnitte das beste Mittel gewihrt solche Fille zu entdecken, so ist 
es ziemlich wahrscheinlich, dass man wirklich diesen Weg eingeschlagen hat.” 
Zeuthen, op. cit. p. 280. 

+ Apollonius of Perga, p. xxv, CXxvi. 

§ Ibid. p. cxxvili and pp. 182, 186 (Conics, v. 58, 62 


ON THE PROBLEMS KNOWN AS ΝΕΎΥΣΕΙΣ. CXV 


to this method as an instance of discovering the solution of a 
‘plane’ problem by means of conics*, the objection having reference 
to the use of a hyperbola where the same points could be obtained 
as the intersections of the parabola with a certain circle. Now the 
proof of this latter fact would present no difficulty to Apollonius, 
and Pappus must have been aware that it would not; if therefore 
he objects in the circumstances to the use of the hyperbola, it is at 
least arguable that he would equally have objected had Apollonius 
brought in the hyperbola and used its properties for the purpose 
of proving the problem to be ‘plane’ in the particular case. 

(2) The solution of the general problem by means of conics 
brings in the auxiliary point P and the straight line CP. We 
should therefore naturally expect to find some trace of these in the 
particular solutions of the νεῦσις for a rhombus and square; but 
they do not appear in the corresponding demonstrations and figures 
given by Pappus. 

Zeuthen considers that the νεῦσις with reference to a square was 
probably shown to be ‘plane’ by means of the same investigation 
which showed that the more general case of the rhombus was also 
capable of solution with the help of the straight line and circle 
only, i.e. by a systematic study of the general solution by means of 
conics. This supposition seems to him more probable than the view 
that the discovery of the plane construction for the square may have 
been accidental ; for (he says) if the same problem is treated solely 
by the aid of elementary geometrical expedients, the discovery that 
it is ‘plane’ is by no means a simple mattery. Here, again, I am 
not convinced by Zeuthen’s argument, as it seems to me that a 
simpler explanation is possible of the way in which the Greeks were 
led to the discovery that the particular νεύσεις were plane. They 
knew in the first place that the trisection of a right angle was a 
‘plane’ problem, and therefore that half a right angle could be 
trisected by means of the straight line and circle. It followed 


* Pappus tv. p. 270. Cf. p. cili above. 

+ ‘Die Ausfiihrbarkeit kann dann auf die zuerst angedeutete Weise gefunden 
sein, die den allgemeinen Fall, wo der Winkel zwischen den gegebenen Geraden 
beliebig ist, in sich begreift. Dies scheint mir viel wahrscheinlicher als die 
Annahme, dass die Entdeckung dieser ebenen Konstruction zufallig sein sollte ; 
denn wenn man dieselbe Aufgabe nur mittels rein elementar-geometrischer 
Hiilfsmittel behandelt, so liegt die Entdeckung, dass sie eben ist, ziemlich fern.” 
Zeuthen, op. cit. p. 282. 


h2 


CXV1 INTRODUCTION. 


therefore that the corresponding νεῦσις, 1.6. that for a square, was 
a ‘plane’ problem in the particular case where the given length 
to which the required intercept was to be equal was double of 
the diagonal of the square. This fact would naturally suggest 
the question whether the problem was still plane if ᾧ had 
any other value; and, when once this question was thoroughly 
investigated, the proof that the problem was ‘plane,’ and the 
solution of it, could hardly have evaded for long the pursuit of 
geometers so ingenious as the Greeks. This will, I think, be 
clear when the solution given by Pappus and reproduced below 
is examined, Again, after it had been proved that the νεῦσις with 
reference to a square was ‘plane,’ what more natural than the further 
inquiry as to whether the intermediate case between that of the 
square and parallelogram, that of the rhombus, might perhaps be a 
‘plane’ problem ? 

As regards the actual solution of the plane νεύσεις with respect 
to the rhombus and square, i.e. the cases in general where the fixed 
point & lies on one of the bisectors of the angles between the two 
given straight lines, Zeuthen says that only in one of the cases have 
we a positive statement that the Greeks solved the νεῦσις by means 
of the circle and ruler, the case, namely, where AC LD is a square*. 
This appears to be a misapprehension, for not only does Pappus 
mention the case of the rhombus as one of the plane νεύσεις which 
the Greeks had solved, but it is clear, from a proposition given by 
him later, how it was actually solved. The proposition is stated 
by Pappus to be “involved” (παραθεωρούμενον, meaning presumably 
“the subject of concurrent investigation”) in the 8th problem of 
Apollonius’ first Book of vetoes, and is enunciated in the following 
formy+. Given a rhombus AD with diameter BC produced to LE, if EF 
be a mean proportional between BE, EC, and if a circle be described 
with centre E and radius EF cutting CD in K and AC produced in 
H, BKH shall be a straight line. The proof is as follows. 

Let the circle cut AC in Z, and join H#, KE, LE. Let LK 
meet BC in 777. 


* “Tndessen besitzen wir doch nur in einem einzelnen hierher gehérigen 
Falle eine positive Angabe dariiber, dass die Griechen die Einschiebung mittels 
Zirkel und Lineal ausgefiihrt haben, wenn nimlich die gegebenen Geraden 
zugleich rechte Winkel bilden, AIBC also ein Quadrat wird.” Zeuthen, op. cit. 
p. 281. 

+ Pappus vit. p. 778. 


ON THE PROBLEMS KNOWN AS NEYSEIS. eCXVll 


Since, from the property of the rhombus, the angles LOM, KCM 
are equal, and therefore CL, Οἱ Καὶ make equal angles with the diameter 
FG of the circle, it follows that CL = CK. 


B D 


Also Hk = EL, and CE is common to the triangles HCK, LCL. 
Therefore the said triangles are equal in all respects, and 
£CKE=LCLE=cCHE. 
Now, by hypothesis, 
EB: EF=EF: EC, 
or EB: EK=EK: EC (since HF = EK), 


and the angle CZK is common to the triangles BEX, KEC; there- 
fore the triangles BEX, KEC are similar, and 


TER OTE me OUD 
= CHE, from above. 


Again, RHCE=LACE— 2 BCK. 

Thus in the triangles CBX, CHE two angles are equal re- 
spectively ; 
therefore i CEH=2CKB. 


But, since . CKE=2 CHE, from above, the points K, C, #, 1 
are concyclic. 
Hence 4 CEH + ἐ CKH = (two right angles). 
Accordingly, since LCEH = CKB, 
. CKH+2CKB = (two right angles), 
and LXH is a straight line. 


CXVill INTRODUCTION. 


Now the form of the proposition at once suggests that, in the 
8th problem referred to, Apollonius had simply given a construction 
involving the drawing of a circle cutting CD and AC produced in 
the points Α΄, H respectively, and Pappus’ proof that BAH is a 
straight line is intended to prove that HK verges towards B, or (in 
other words) to verify that the construction given by Apollonius 
solves a certain νεῦσις requiring BKH to be drawn so that KH ἐξ 
equal to a given length. 

The analysis leading to the construction must have been worked 
out somewhat as follows. 

Suppose ΒΑ drawn so that AH is equal to the given length ὦ. 
Bisect KH at NV, and draw NE at right angles to KH meeting BC 
produced in ἢ. 

Draw KW perpendicular to BC and produce it to meet CA in ὦ. 
Then, from the property of the rhombus, the triangles KCM, LCM 
are equal in all respects. 

Therefore KM= ML; and accordingly, if IN be joined, JN, 
LH are parallel. 

Now, since the angles at 1, WV are right, a circle can be described 
about ΚΑ. 


Therefore .CEK=2zMNK&, in the same segment, 
= CHK, by parallels. 
Hence a circle can be described about CHHK. It follows that 
LBCD=.CEK+LCKE 
=.CHK+LCHE 
= 2 oe 
Therefore the triangles HAH, DBC are similar. 
Lastly, LCKN=LCBK+2BCK; 


and, subtracting from these equals the equal angles HAN, BCK 


respectively, we have 
LEKC =c EBK. 


Hence the triangles HBX, FKC are similar, and 
BE: HR=EK : EC, 
or BE. EC = Ek’. 
But, by similar triangles, 7K: KH = DC: CB, 
and the ratio DC : CB is given, while KH is also given (= 4). 


ON THE PROBLEMS KNOWN AS NETZEI®, CX1X 


Therefore HX is given, and, in order to find #, we have only, in 
the Greek phrase, to ‘‘apply to BC a rectangle exceeding by a square 
figure and equal to the given area HK.” 

Thus the construction given by Apollonius was clearly the 


following*. 
If k be the given length, take a straight line p such that 


pik AB BC. 


Apply to BC a rectangle exceeding by a square figure and equal to 
the area p’. Let BE. EC be this rectangle, and with E as centre and 
radius equal to p describe a circle cutting AC produced in H and 


CD in K. 
HK is then equal to ἄ, and verges towards Bb, as proved by 
Pappus; the problem is therefore solved. 


The construction used by Apollonius for the ‘plane’ νεῦσις with 
reference to the rhombus having been thus restored by means of the 
theorem given by Pappus, we are enabled to understand the purpose 


* This construction was suggested to me by a careful examination of 
Pappus’ proposition without other aid; but it is no new discovery. 
Samuel Horsley gives the same construction in his restoration of Apollonii 
Pergaei Inclinationum libri duo (Oxford, 1770); he explains, however, that 
he went astray in consequence of a mistake in the figure given in the mss., 
and was unable to deduce the construction from Pappus’s proposition until he 
was recalled to the right track by a solution of the same problem by Hugo 
@Omerique. This solution appears in a work entitled, Analysis geometrica, sive 
nova et vera methodus resolvendi tam problemata geometrica quam arithmeticas 
quaestiones, published at Cadiz in 1698. D’Omerique’s construction, which is 
practically identical with that of Apollonius, appears to have been evolved by 
means of an independent analysis of his own, since he makes no reference to 
Pappus, as he does in other cases where Pappus is drawn upon (e.g. when giving 
the construction for the case of the square attributed by Pappus to one 
Heraclitus). The construction differs from that given above only in the fact 
that the circle is merely used to determine the point K, after which BK is joined 
and produced to meet AC in H. Of other solutions of the same problem two 
may here be mentioned. (1) The solution contained in Marino Ghetaldi’s 
posthumous work De Resolutione et Compositione Mathematica Libri quinque 
(Rome, 1630), and included among the solutions of other problems all purporting 
to be solved ‘‘ methodo qua antiqui utebantur,” is, though geometrical, entirely 
different from that above given, being effected by means of a reduction of the 
problem to a simpler plane νεῦσις of the same character as that assumed by 
Hippocrates in his Quadrature of lunes. (2) Christian Huygens (De circuli 
magnitudine inventa; accedunt problematum quorundam illustrium constructiones, 
Lugduni Batavorum, 1654) gave a rather complicated solution, which may be 
described as a generalisation of Heraclitus’ solution in the case of a square. 


CXX INTRODUCTION. 


for which Pappus, while still on the subject of the “8th problem ” 
of Apollonius, adds a solution for the particular case of the square 
(which he calls a ‘‘ problem after Heraclitus”) with an introductory 
lemma. It seems clear that Apollonius did not treat the case of the 
square separately from the rhombus because the solution for the 
rhombus was equally applicable to the square, and this supposition 
is confirmed by the fact that, in setting out the main problems 
discussed in the νεύσεις, Pappus only mentions the rhombus and not 
the square. Being however acquainted with a solution by one 
Heraclitus of the νεῦσις relating to a square which was not on the 
same lines as that of Apollonius, while it was not applicable to the 
case of the rhombus, Pappus adds it as an alternative method for 
the square which is worth noting*. This is no doubt the explanation 
of the heading to the lemma prefixed to Heraclitus’ problem which 
Hultsch found so much difficulty in explaining and put in brackets 
as an interpolation by a writer who misunderstood the figure 
and the object of the theorem. The words mean ‘“‘ Lemma useful 
for the [problem] with reference to squares taking the place 
of the rhombus” (literally “having the same property as the 
rhombus”), 1.6. a lemma useful for Heraclitus’ solution of the 


* This view of the matter receives strong support from the following 
facts. In Pappus’ summary (p. 670) of the contents of the νεύσεις of Apollonius 
‘two cases” of the νεῦσις with reference to the rhombus are mentioned last 
among the particular problems given in the first of the two Books. As we have 
seen, one case (that given above) was the subject of the ‘‘8th problem” of 
Apollonius, and it is equally clear that the other case was dealt with in the 
“9th problem.” The other case is clearly that in which 
the line to be drawn through B, instead of crossing the ἢ A [Ὁ 
exterior angle of the rhombus at C, lies across the angle 
C itself, i.e. meets CA, CD both produced. In the former 
case the solution of the problem is always possible what- 
ever be the length of k; but in the second case clearly 
the problem is not capable of solution if k, the given 
length, is less than a certain minimum. Hence the 
problem requires a διορισμός to determine the minimum K 
length of k. Accordingly we find Pappus giving, after 
the interposition of the case of the square, a ‘‘lemma useful for the διορισμός of 
the 9th problem,” which proves that, if CH=CK and B be the middle point of 
HK, then HK is the least straight line which can be drawn through B to meet 
CH, CK. Pappus adds that the διορισμός for the rhombus is then evident; if 
HK be the line drawn through B perpendicular to CB and meeting CA, CD 
produced in H, K, then, in order that the problem may admit of solution, the 
given length k must be not less than HK. 


ON THE PROBLEMS KNOWN AS ΝΕΥΣΕΙΣ. CXX1 


νεῦσις in the particular case of a square*. The lemma is as 
follows. 

ABCD being a square, suppose BHE drawn so as to meet CD in 
H and AD produced in E, and let EF be drawn perpendicular to BE 
meeting BC produced in F. To prove that 

CF’ = BC? + ΜΕ". 

Suppose HG drawn parallel to DC meeting CF in G. Then 

since BEF is a right angle, the angles HBC, FEE are equal. 


hee ae ae 
B 


ἜΣ ΠΩΣ 


c G F 


Therefore the triangles BCH, HGF are equal in all respects, and 


EF=BH. 
Now BF? = BE? + ΕΓ", 
or BC. BF+BF.FC=BH. BE +BE.EH + EF". 


But, the angles HCF, HEF being right, the points C, H, Z, F 
are concyclic, and therefore 
BC, BY = BH, BH. 
Subtracting these equals, we have 
BF. FC=BE.EH+ EF?’ 
= BE. EH + BH’ 
= BH. HE + EH’ + BH’ 
=HB.BH + HH? 
= FB. BC+ EF’. 


* Hultsch translates the words λῆμμα χρήσιμον eis τὸ ἐπὶ τετραγώνων ποιούντων 
τὰ αὐτὰ τῷ ῥόμβῳ (p. 780) thus, “‘ Lemma utile ad problema de quadratis quorum 
summa rhombo aequalis est,” and has a note in his Appendix (p. 1260) explaining 
what he supposes to be meant. The ‘squares’ he takes to be the given square 
and the square on the given length of the intercept, and the rhombus to be one 
for which he indicates a construction but which is not shown in Pappus’ figure. 
Thus he is obliged to translate τῷ ῥόμβῳ as ‘a rhombus,” which is one objec- 
tion to his interpretation, while “whose squares are equal” scarcely seems a 
possible rendering of ποιούντων τὰ αὐτά. 


ΟΧΧΙΙ INTRODUCTION. 


Take away the common part BC. CF, and 
Cl? = BC + LR", 
Heraclitus’ analysis and construction are now as follows. 
Suppose that we have drawn BHE so that HE has a given 
length &. 


Since CF* = BC’? + EH’, or BC? + k’, 
and BC and k are both given, 
CF is given, and therefore BF is given. 


Thus the semicircle on BF as diameter is given, and therefore 
also #, its intersection with the given line ADE; hence BE is 
given. 

To effect the construction, we first find a square equal to the 
sum of the given square and the square on &. We then produce 
BC to F so that CF is equal to the side of the square so found. If 
a semicircle be now described on BF as diameter, it will pass above 
D (since CF'> CD, and therefore BC. CF > CD”), and will therefore 
meet AD produced in some point £. 

Join BE meeting CD in H. 

Then HE =k, and the problem is solved. 


CHAPTER VI. 


CUBIC EQUATIONS. 


Ir has often been explained how the Greek geometers were able 
to solve geometrically all forms of the quadratic equation which give 
positive roots; while they could take no account of others because 
the conception of a negative quantity was unknown to them. The 
quadratic equation was regarded as a simple equation connecting 
areas, and its geometrical expression was facilitated by the methods 
which they possessed of transforming any rectilineal areas whatever 
into parallelograms, rectangles, and ultimately squares, of equal 
area ; its solution then depended on the principle of application of 
areas, the discovery of which is attributed to the Pythagoreans. 
Thus any plane problem which could be reduced to the geometrical 
equivalent of a quadratic equation with a positive root was at once 
solved. <A particular form of the equation was the pure quadratic, 
which meant for the Greeks the problem of finding a square equal 
to a given rectilineal area. This area could be transformed into a 
rectangle, and the general form of the equation thus became a? = ab, 
so that it was only necessary to find a mean proportional between ὦ 
and ὁ. In the particular case where the area was given as the 
sum of two or more squares, or as the difference of two squares, 
an alternative method depended on the Pythagorean theorem of 
Eucl. 1. 47 (applied, if necessary, any number of times successively). 
The connexion between the two methods is seen by comparing 
Eucl. vi. 13, where the mean proportional between a and ὦ is 
found, and Eucl. τι. 14, where the same problem is solved without 
the use of proportions by means of 1. 47, and where in fact the 


formula used is 
ae GONG a—b\2 
ἘΠῚ ΞΞ - (4) -( 5) ) A 


CXX1V INTRODUCTION. 


The choice between the two methods was equally patent when the 
equation to be solved was «*=pa*, where p is any integer; hence 
the ‘multiplication’ of squares was seen to be dependent on the 
finding of a mean proportional. The equation x= 2a’ was the 
simplest equation of the kind, and the discovery of a geometrical 
construction for the side of a square equal to twice a given square 
was specially important, as it was the beginning of the theory of 
incommensurables or ‘irrationals’ (ἀλόγων πραγματεία) which was 
invented by Pythagoras. There is every reason to believe that this 
successful doubling of the square was what suggested the question 
whether a construction could not be found for the doubling of the 
cube, and the stories of the tomb erected by Minos for his son and 
of the oracle bidding the Delians to double a cubical altar were no 
doubt intended to invest the purely mathematical problem with an 
element of romance. It may then have been the connexion between 
the doubling of the square and the finding of one mean proportional 
which suggested the reduction of the doubling of the cube to the 
problem of finding two mean proportionals between two unequal 
straight lines. This reduction, attributed to Hippocrates of Chios, 
showed at the same time the possibility of multiplying the cube 
by any ratio. Thus, if «, y are two mean proportionals between 
a, ὦ, we have 
τ ΞΘ τ ya sD. 


and we derive at once 


G20 es 


whence a cube (2*) is obtained which bears to οὐ the ratio ὦ : a, 
P 


while any fraction ~ can be transformed into a ratio between lines 
qY 


of which one (the consequent) is equal to the side a of the given 
cube. Thus the finding of two mean proportionals gives the solution 
of any pure cubic equation, or the equivalent of extracting the cube 
root, just as the single mean proportional is equivalent to extracting 
the square root. For suppose the given equation to be 2’ = bed. 
We have then only to tind a mean proportional ὦ between ὁ and d, 


: ᾿ ὐ ΝΕ 
and the equation becomes #*=a?.b=a*.— which is exactly the 
a 


multiplication of a cube by a ratio between lines which the two 
mean proportionals enable us to effect. 

As a matter of fact, we do not find that the great geometers 
were in the habit of reducing problems to the multiplication of the 


CUBIC EQUATIONS. CXXV 


cube eo nomine, but to the equivalent problem of the two mean 
proportionals ; and the cubic equation x* -- αὐὖ is not usually stated 
in that form but as a proportion. Thus in the two propositions On 
the Sphere and Cylinder τι. 1, 5, where Archimedes uses the two 
mean proportionals, it is required to find « where 
(op Hp ane 

he does not speak of finding the side of a cube equal to a certain 
parallelepiped, as the analogy of finding a square equal to a given 
rectangle might have suggested. So far therefore we do not find 
any evidence of a general system of adding and subtracting solids 
by transforming parallelepipeds into cubes and cubes into parallel- 
epipeds which we should have expected to see in operation if the 
Greeks had systematically investigated the solution of the general 
form of the cubic equation by a method analogous to that of the 
application of areas employed in dealing with quadratic equations. 

The question then arises, did the Greek geometers deal thus 
generally with the cubic equation 

αὐ τ ααὐ-π Be+T=0, 

which, on the supposition that it was regarded as an independent 
problem in solid geometry, would be for them a simple equation 
between solid figures, ὦ and a both representing linear magnitudes, 
B an area (a rectangle), and I a volume (a parallelepiped)? And 
was the reduction of a problem of an order higher than that which 
could be solved by means of a quadratic equation to the solution of 
a cubic equation in the form shown above a regular and recognised 
method of dealing with such a problem? The only direct evidence 
pointing to such a supposition is found in Archimedes, who reduces 
the problem of dividing a sphere by a plane into two segments 
whose volumes are in a given ratio (On the Sphere and Cylinder τι. 4) 
to the solution of a cubic equation which he states in a form 
equivalent to 


where ὦ is the radius of the sphere, m:n the given ratio (being a 
ratio between straight lines of which m>n), and x the height of the 
greater of the required segments. Archimedes explains that this is 
a particular case of a more general problem, to divide a straight 
line (a) into two parts (ὦ, a—«) such that one part (a—«) is to an- 
other given straight line (c) as a given area (which for convenience’ 


ΟΧΧΝῚ INTRODUCTION, 


sake we suppose transformed into a square, ὦ) is to the square on 
the other part («’), i.e. so that 


Gudea πε σεις" Βρω |) 


He further explains that the equation (2) stated thus generally 
requires a διορισμός, i.e. that the limits for the possibility of a real 
solution, etc., require to be investigated, but that the particular case 
(with the conditions obtaining in the particular proposition) requires 
no διορισμός, i.e. the equation (1) will always give a real solution. 
He adds that ‘“‘the analysis and synthesis of both these problems 
will be given at the end.” That is, he promises to give separately a 
complete investigation of the equation (2), which is equivalent to the 
cubic equation 


and to apply it to the particular case (1). 

Wherever the solution was given, it was temporarily lost, having 
apparently disappeared even before the time of Dionysodorus and 
Diocles (the latter of whom lived, according to Cantor, not later 
than about 100 B.c.); but Eutocius describes how he found an 
old fragment which appeared to contain the original solution of 
Archimedes, and gives it in full. It will be seen on reference to 
Eutocius’ note (which I have reproduced immediately after the 
proposition to which it relates, On the Sphere and Cylinder τι. 4) 
that the solution (the genuineness of which there seems to be no 
reason to doubt) was effected by means of the intersection of a 
parabola and a rectangular hyperbola whose equations may re- 
spectively be written thus, 


ΤΉ τ 
w= τ 
(ὦ -- α) y=ac, 


The διορισμός takes the form of investigating the maximum 
possible value of x? (a—2), and it is proved that this maximum 


2 
value for a real solution is that corresponding to the value x= 5 a. 


3 
This is established by showing that, if b’c == a’, the curves touch 
2 
at the point for which «= 3 ὦ. If on the other hand δ᾽ < = a’, it 


is proved that there are two real solutions. In the particular case 
(1) it is clear that the condition for a real solution is satisfied, for 


CUBIC EQUATIONS. CXxVil 


ne 


the expression in (1) corresponding to 6’c in (2) is i 4a’, and it 
is only necessary that 


m 


4a? > = (3a)*, or 4%, 


m+n 

which is obviously true. 

Hence it is clear that not only did Archimedes solve the cubic 
equation (3) by means of the intersections of two conics, but he also 
discussed completely the conditions under which there are 0, 1 or 2 
roots lying between 0 and a. It is to be noted further that the 
διορισμός is similar in character to that by which Apollonius 
investigates the number of possible normals that can be drawn 
to a conic from a given point*. Lastly, Archimedes’ method is 
seen to be an extension of that used by Menaechmus for the solution 
of the pure cubic equation. This can be put in the form 

ὩΣ ΞΘ Ds 
which can again be put in Archimedes’ form thus, 
Ce τ ἢ 
and the conics used by Menaechmus are respectively 
x = ay, cy = ab, 
which were of course suggested by the two mean proportionals 
satisfying the equations 
Ge ΞΡ τυ ΞΡ: 0 

The case above described is not the only one where we may 
assume Archimedes to have solved a problem by first reducing it 
to a cubic equation and then solving that. At the end of the 
preface to the book On Conoids and Spheroids he says that the 
results therein obtained may be used for discovering many theorems 
and problems, and, as instances of the latter, he mentions the 
following, “from a given spheroidal figure or conoid to cut off, 
by a plane drawn parallel to a given plane, a segment which shall 
be equal to a given cone or cylinder, or to a given sphere.” Though 
Archimedes does not give the solutions, the following considerations 
may satisfy us as to his method. 


(1) The case of the ‘right-angled conoid’ (the paraboloid of 
revolution) is a ‘plane’ problem and therefore does not concern us 
here. 


* Cf. Apollonius of Perga, p. 168 sqq. 


CXXVill INTRODUCTION. 


(2) In the case of the spheroid, the volume of the whole 
spheroid could be easily ascertained, and, by means of that, the 
ratio between the required segment and the remaining segment ; 
after which the problem could be solved in exactly the same way 
as the similar one in the case of the sphere above described, 
since the results in On Conoids and Spheroids, Props. 29—32, 
correspond to those of On the Sphere and Cylinder τι. 2. Or 
Archimedes may have proceeded in this case by a more direct 
method, which we may represent thus. Let a plane be drawa 
through the axis of the spheroid perpendicular to the given 
plane (and therefore to the base of the required segment). This 
plane will cut the elliptical base of the segment in one of its 
axes, which we will call 2y. Let α be the length of the axis 
of the segment (or the length intercepted within the segment 
of the diameter of the spheroid passing through the centre of the 
base of the segment). Then the area of the base of the segment will 
vary as y” (since all sections of the spheroid parallel to the given 
plane must be similar), and therefore the volume of the cone which 
has the same vertex and base as the required segment will vary as 
yx. And the ratio of the volume of the segment to that of the 
cone is (On Conoids and Spheroids, Props. 29—32) the ratio 
(3a — a): (2a—a), where 2a is the length of the diameter of the 
spheroid which passes through the vertex of the segment. There- 
fore 

96 — ὦ 


Yn . = 
Maia Pee : 


where C is a known volume. Further, since x, y are the coordinates 
of a point on the elliptical section of the spheroid made by the plane 
through the axis perpendicular to the cutting plane, referred to a 
diameter of that ellipse and the tangent at the extremity of the 
diameter, the ratio γῆ: ὦ (3α -- α) is given. Hence the equation 
can be put in the form 
a? (3a — 2) = b*e, 

and this again is the same equation as that solved in the fragment 
given by Eutocius. A διορισμός is formally necessary in this case, 
though it only requires the constants to be such that the volume 
to which the segment is to be equal must be less than that of the 
whole spheroid. 


(3) For the ‘obtuse-angled conoid’ (hyperboloid of revolution) 
it would be necessary to use the direct method just described for 


CUBIC EQUATIONS. ΟΧΧΙΧ 


the spheroid, and, if the notation be the same, the corresponding 
equations will be found, with the help of On Conoids and Spheroids, 
Props. 25, 26, to be 

eas 


7 
γα. = 
“ ϑατα 


? 


and, since the ratio 7” : a (2a + 2) is constant, 
αὐ (3a + x) = b’e. 


Tf this equation is written in the form of a proportion like the 
similar one above, it becomes 
ὅν Ὁ = (Sa'+ia).2 δ. 

There can be no doubt that Archimedes solved this equation as 
well as the similar one with a negative sign, i.e. he solved the two 
equations 

αὐ + au’ ¥ b’c=0, 
obtaining all their positive real roots. In other words, he solved 
completely, so far as the real roots are concerned, a cubic equation 
in which the term in α is absent, although the determination of the 
positive and negative roots of one and the same equation meant for 
him two separate problems. And it is clear that all cubic equations 
can be easily reduced to the type which Archimedes solved. 

We possess one other solution of the cubic equation to which 
the division of a sphere into segments bearing a given ratio to one 
another is reduced by Archimedes. This solution is by Dionysodorus, 
and is given in the same note of Eutocius*. Dionysodorus does not 
generalise the equation, however, as is done in the fragment quoted 
above ; he merely addresses himself to the particular case, 


4a’: x? = (Ξα-- αν : Ξ ΤῊ 26 


m+n’ 


thereby avoiding the necessity for a διορισμός. The curves which he 
uses are the parabola 


m ire 
m+n el 
and the rectangular hyperbola 
m 
—— 2a? = xy. 
m+n r Ἵ 


When we turn to Apollonius, we find him emphasising in his 


* On the Sphere and Cylinder τι. 4 (note at end). 


CXXX INTRODUCTION, 


preface to Book tv. of the Conics* the usefulness of investigations 
of the possible number of points in which conics may intersect one 
another or circles, because “they at all events afford a more ready 
means of observing some things, e.g. that several solutions are 
possible, or that they are so many in number, and again that no 
solution is possible”; and he shows his mastery of this method 
of investigation in Book v., where he determines the number of 
normals that can be drawn to a conic through any given point, the 
condition that two normals through it coincide, or (in other words) 
that the point lies on the evolute of the conic, and so on. For these 
purposes he uses the points of intersection of a certain rectangular 
hyperbola with the conic in question, and among the cases we find 
(v. 51, 58, 62) some which can be reduced to cubic equations, those 
namely in which the conic is a parabola and the axis of the parabola 
is parallel to one of the asymptotes of the hyperbola. Apollonius 
however does not bring in the cubic equation ; he addresses himself 
to the direct geometrical solution of the problem in hand without 
reducing it to another. This is after all only natural, because the 
solution necessitated the drawing of the rectangular hyperbola in 
the actual figure containing the conic in question ; thus, e.g. in the 
case of the problem leading to a cubic equation, Apollonius can, so 
to speak, compress two steps into one, and the introduction of the 
cubic as such would be mere surplusage. The case was different 
with Archimedes, when he had no conic in his original figure ; and 
the fact that he set himself to solve a cubic somewhat more general 
than that actually involved in the problem made separate treatment 
with a number of new figures necessary. Moreover Apollonius was 
at the same time dealing, in other propositions, with cases which did 
not reduce to cubics, but would, if put in an algebraical form, lead 
to biquadratic equations, and these, expressed as such, would have 
had no meaning for the Greeks ; there was therefore the less reason 
in the simpler case to introduce a subsidiary problem. 

As already indicated, the cubic equation, as a subject of syste- 
matic and independent study, appears to have been lost sight of 
within a century or so after the death of Archimedes. Thus Diocles, 
the discoverer of the cissoid, speaks of the problem of the division of 
the sphere into segments in a given ratio as having been reduced 
by Archimedes ‘to another problem, which he does not solve in 
his work on the sphere and cylinder”; and he then proceeds to 


* Apollonius of Perga, p. 1xxiii. 


CUBIC EQUATIONS. CXXX1 


solve the original problem directly, without in any way bringing 
in the cubic. This circumstance does not argue any want of 
geometrical ability in Diocles; on the contrary, his solution of the 
original problem is a remarkable instance of dexterity in the use of 
conics for the solution of a somewhat complicated problem, and it 
proceeds on independent lines in that it depends on the intersection 
of an ellipse and a rectangular hyperbola, whereas the solutions of 
the cubic equation have accustomed us to the use of the parabola 
and the rectangular hyperbola. I have reproduced Diocles’ solution 
in its proper place as part of the note of Eutocius on Archimedes’ 
proposition ; but it will, I think, be convenient to give here its 
equivalent in the ordinary notation of analytical geometry, in 
accordance with the plan of this chapter. Archimedes had proved 
[On the Sphere and Cylinder τι. 2] that, if & be the height of a 
segment cut off by a plane from a sphere of radius a, and if h be 
the height of the cone standing on the same base as that of the 
segment and equal in volume to the segment, then 


(ϑα -- ἀ) : (Qa—-—k)=h : 1. 


Also, if h’ be the height of the cone similarly related to the 
remaining segment of the sphere, 


(a+k):k=h': (2a—h). 
From these equations we derive 

(h—k):k=a: (2a—h), 
and (h'—2a+k): (2a --ἀγ τεῳ: ἢ. 

Slightly generalising these equations by substituting for ὦ in the 
third term of each proportion another length ὦ, and adding the 
condition that the segments (and therefore the cones) are to bear to 
each other the ratio m:n, Diocles sets himself to solve the three 
equations 

(h—k):k=6: (2a—hk) 
(h’—Qa+k):(2Q4—h)=b:b& | ceeveevsesee νον (A). 


and A:h’=m:n 


Suppose m>n, so that k>a. The problem then is to divide a 
straight line of length 2a into two parts k and (2a—h) of which & is 
the greater, and which are such that the three given equations are 
all simultaneously satisfied. 

Imagine two coordinate axes such that the origin is the middle 
point of the given straight line, the axis of y is at right angles to it, 

42 


“ἰ 


exxxil INTRODUCTION. 


and a is positive when measured along that half of the given straight 
line which is to contain the required point of division. Then the 
conics drawn by Diocles are 


(1) the ellipse represented by the equation 
(y+a-2)*=— {(a+ ba, 


and (2) the rectangular hyperbola 

(a + a) (y + 6) = 2ab. 
One intersection between these conics gives a value of x between 0 
and a, and leads to the solution required. Treating the equations 
algebraically, and eliminating y by means of the second equation 
which gives 


we obtain from the first equation 


(a—x)? (: ἡγοῦ ᾿Ξ 5. sq +b) ὧι, 


a+ x m 


that is, (a+ a)? (α -- ὃ -- α) ΞΞ ~ (ὦ -- 2)" (ὦ Ὁ 542) .τ. {τς (B). 


In other words Diocles’ method is the equivalent of solving a 
complete cubic equation containing all the three powers of α and a 
constant, though no mention is made of such an equation. 

To verify the correctness of the result we have only to remember 
that, « being the distance of the point of division from the middle 
point of the given straight line, 


k=a+a, 2a-k=a-x. 
Thus, from the first two of the given equations (A) we obtain 
respectively 


᾿ a+x 

h=a+x+——.6, 
α--α 

h'=a-2+—.}, 
a+a 


whence, by means of the third equation, we derive 
(a+2)'(a+b—-a)=™ (a-2) (a +b+«), 


which is the same equation as that found by elimination above (B). 


CUBIC EQUATIONS. CXXXill 


I have purposely postponed, until the evidence respecting the 
Greek treatment of the cubic equation was complete, any allusion 
to an interesting hypothesis of Zeuthen’s* which, if it could be 
accepted as proved, would explain some difficulties involved in 
Pappus’ account of the orthodox classification of problems and loci. 
I have already quoted the passage in which Pappus distinguishes 
the problems which are plane (ἐπίπεδα), those which are solid (στερεά) 
and those which are linear (ypappuxa)}. Parallel to this division of 
problems into three orders or classes is the distinction between three 
classes of /oci{. The first class consists of plane loci (τόποι ἐπίπεδοι) 
which are exclusively straight lines and circles, the second of solid 
loct (τόποι στερεοί) which are conic sections§, and the third of 
linear loci (τόποι γραμμικοί). It is at the same time clearly implied 
by Pappus that problems were originally called plane, solid or linear 
respectively for the specific reason that they required for their 
solution the geometrical loci which bore the corresponding names. 
But there are some logical defects in the classification both as 
regards the problems and the loci. 


(1) Pappus speaks of its being a serious error on the part of 
geometers to solve a plane problem by means of conics (i.e. ‘solid 
loci’) or ‘linear’ curves, and generally to solve a problem “ by means 
of a foreign kind” (ἐξ ἀνοικείου γένους). If this principle were 
applied strictly, the objection would surely apply equally to the 
solution of a ‘solid’ problem by means of a ‘linear’ curve. Yet, 
though e.g. Pappus mentions the conchoid and the cissoid as being 
‘linear’ curves, he does not object to their employment in the 
solution of the problem of the two mean proportionals, which is a 
‘solid’ problem. 


(2) The application of the term ‘solid loci’ to the three conic 
sections must have reference simply to the definition of the curves 
as sections of a solid figure, viz. the cone, and it was no doubt in 
contrast to the ‘solid locus’ that the ‘plane locus’ was so called. 
This agrees with the statement of Pappus that ‘ plane’ problems may 


* Die Lehre von den Kegelschnitten, p. 226 sqq. 

+ Be Ciil. 

+ Pappus vit. pp. 652, 662. 

8 It is true that Proclus (p. 394, ed. Friedlein) gives a wider definition of 
‘‘ solid lines” as those which arise ‘‘ from some section of a solid figure, as the 
cylindrical helix and the conic curves’’; but the reference to the cylindrical 
helix would seem to be due to some confusion. 


CXXX1V INTRODUCTION. 


properly be so called because the lines by means of which they are 
solved “have their origin in a plane.” But, though this may be 
regarded as a satisfactory distinction when ‘plane’ and ‘solid’ loci 
are merely considered in relation to one another, it becomes at once 
logically defective when the third or ‘linear’ class is also brought 
in. For, on the one hand, Pappus shows how the ‘ quadratrix’ (a 
‘linear’ curve) can be produced by a construction in three 
dimensions (“by means of surface-loci,” διὰ τῶν πρὸς ἐπιφανείαις 
τόπων) ; and, on the other hand, other ‘linear’ loci, the conchoid 
and cissoid, have their origin in a plane. If then Pappus’ account 
of the origin of the terms ‘plane’ and ‘solid’ as applied to problems 
and loci is literally correct, it would seem necessary to assume that 
the third name of ‘linear’ problems and loci was not invented until 
a period when the terms ‘plane’ and ‘solid loci’ had been so long 
recognised and used that their origin was forgotten. 

To get rid of these difficulties, Zeuthen suggests that the terms 
‘plane’ and ‘solid’ were first applied to problems, and that they 
came afterwards to be applied to the geometrical loci which were 
used for the purpose of solving them. On this interpretation, when 
problems which could be solved by means of the straight line and 
circle were called ‘plane,’ the term is supposed to have had reference, 
not to any particular property of the straight line or circle, but to 
the fact that the problems were such as depend on an equation of a 
degree not higher than the second. The solution of a quadratic 
equation took the geometrical form of application of areas, and the 
term ‘plane’ became a natural one to apply to the class of problems 
so soon as the Greeks found themselves confronted with a new class 
of problems to which, in contrast, the term ‘solid’ could be applied. 
This would happen when the operations by which problems were 
reduced to applications of areas were tried upon problems which 
depend on the solution of a cubic equation. Zeuthen, then, 
supposes that the Greeks sought to give this equation a similar 
shape to that which the reduced ‘plane’ problem took, that is, to 
form a simple equation between solids corresponding to the cubic 
equation 

αὐ +ae+ Bu+T=0; 
the term ‘solid’ or ‘plane’ being then applied according as it had 
been reduced, in the manner indicated, to the geometrical equivalent 
of a cubic or a quadratic equation. 

Zeuthen further explains the term ‘linear problem’ as having 


CUBIC EQUATIONS. CXXXV 


been invented afterwards to describe the cases which, being 
equivalent to algebraical equations of an order higher than the 
third, would not admit of reduction to a simple relation between 
lengths, areas and volumes, and either could not be reduced to an 
equation at all or could only be represented as such by the use of 
compound ratios. The term ‘linear’ may perhaps have been applied 
because, in such cases, recourse was had to new classes of curves, 
directly and without any intermediate step in the shape of an 
equation. Or, possibly, the term may not have been used at all 
until a time when the original source of the names ‘plane’ and 
‘solid’ problems had been forgotten. 

On these assumptions, it would still be necessary to explain how 
Pappus came to give a more extended meaning to the term ‘solid 
problem,’ which according to him equally includes those problems 
which, though solved by the same method of conics as was used to 
solve the equivalent of cubics, do not reduce to cubic equations but 
to biquadratics. This is explained by the supposition that, the 
cubic equation having by the time of Apollonius been obscured 
from view owing to the attention given to the method of solution 
by means of conics and the discovery that the latter method was 
one admitting of wider application, the possibility of solution by 
means of conics came itself to be regarded as the criterion deter- 
mining the class of problem, and the name ‘solid problem’ came 
to be used in the sense given to it by Pappus through a natural 
misapprehension. A similar supposition would account, in Zeuthen’s 
view, for a circumstance which would otherwise seem strange, viz. 
that Apollonius does not use the expression ‘solid problem,’ though 
it might have been looked for in the preface to the fourth Book 
of the Conics. The term may have been avoided by Apollonius 
because it then had the more restricted meaning attributed to it by 
Zeuthen and therefore would not have been applicable to all the 
problems which Apollonius had in view. 

It must be admitted that Zeuthen’s hypothesis is in several 
respects attractive. I cannot however feel satisfied that the 
positive evidence in favour of it is sufficiently strong to outweigh 
the authority of Pappus where his statements tell the other way. 
To make the position clear, we have to remember that Menaechmus, 
the discoverer of the conic sections, was a pupil of Eudoxus who 
flourished about 365 B.c.; probably therefore we may place the 
discovery of conics at about 350 B.c. Now Aristaeus ‘the elder’ 


CXXXV1 INTRODUCTION. 


wrote a book on solid loci (στερεοὶ τόποι) the date of which Cantor 
concludes to have been about 320 B.c. Thus, on Zeuthen’s hypo- 
thesis, the ‘solid problems’ the solution of which by means of conics 
caused the latter to be called ‘solid loci’ must have been such as 
had been already investigated and recognised as solid problems 
before 320 B.c., while the definite appropriation, so to speak, of the 
newly discovered curves to the service of the class of problems must 
have come about in the short period between their discovery and 
the date of Aristaeus’ work. It is therefore important to consider 
what particular problems leading to cubic equations appear to have 
been the subject of speculation before 320 B.c. We have certainly 
no ground for assuming that the cubic equation used by Archimedes 
(On the Sphere and Cylinder τι. 4) was one of these problems ; for 
the problem of cutting a sphere into segments bearing a given ratio 
to one another could not have been investigated by geometers who 
had not succeeded in finding the volume of a sphere and a segment 
of a sphere, and we know that Archimedes was the first to discover 
this. On the other hand there was the duplication of the cube, or 
the solution of a pure cubic equation, which was a problem dating 
from very early times. Also it is certain that the trisection of an 
angle had Jong exercised the minds of the Greek geometers. Pappus 
says that “the ancient geometers” considered this problem and first 
tried to solve it, though it was by nature a solid problem (πρόβλημα 
τῇ φύσει στερεὸν ὑπάρχον), by means of plane considerations (διὰ τών 
ἐπιπέδων) but failed; and we know that Hippias of Elis invented, 
about 420 B.c,, a transcendental curve which was capable of being 
used for two purposes, the trisection of an angle, and the quadrature 
of a circle*. This curve came to be called the Quadratrixt, but, as 
Deinostratus, a brother of Menaechmus, was apparently the first to 
apply the curve to the quadrature of the circle{, we may no doubt. 
conclude that it was originally intended for the purpose of trisecting 


* Proclus (ed. Friedlein), p. 272. 

+ The character of the curve may be described as follows. Suppose there 
are two rectangular axes Oy, Ox and that a straight line OP of a certain length 
(a) revolves uniformly from a position along Oy to a position along Ox, while a 
straight line remaining always parallel to Ox and passing through P in its 
original position also moves uniformly and reaches Ox in the same time as the 
moving radius OP. The point of intersection of this line and OP describes the 
Quadratrix, which may therefore be represented by the equation 

y/a=20/7. 
+ Pappus tv. pp. 250—2. 


CUBIC EQUATIONS. CXXXV1l 


an angle. Seeing therefore that the Greek geometers had used their 
best efforts to solve this problem before the invention of conics, it 
may easily be that they had succeeded in reducing it to the geo- 
metrical equivalent of a cubic equation. They would not have been 
unequal to effecting this reduction by means of the figure of the 
νεῦσις given above on p. ΟΧΙ]. with a few lines added. The proof 
would of course be the equivalent of eliminating « between the two 
equations 


ay = ab 

αὐ ον eae, ἜΤΙ een (a) 
where «= DF, y=FP= EC, a=DA, b= DB. 
The second equation gives 

(a + a) (ὦ -- 3a) = (y +b) (3b -- ψ). 
From the first equation it is easily seen that 

(c+a):(y+b)=a:y, 

and that (x — 3a) y=a(b — 3y) ; 
we have therefore GO 3) a OD — Of) eens ere -- (8) 
[or γ᾽ — 3by? — 3a*y + ab = ΟἹ. 


If then the trisection of an angle had been reduced to the geo- 
metrical equivalent of this cubic equation, it would be natural for 
the Greeks to speak of it as a solid problem. In this respect it 
would be seen to be similar in character to the simpler problem of 
the duplication of the cube or the equivalent of a pure cubic 
equation; and it would be natural to see whether the transformation 
of volumes would enable the mixed cubic to be reduced to the form 
of the pure cubic, in the same way as the transformation of areas 
enabled the mixed quadratic to be reduced to the pure quadratic. 
The reduction to the pure cubic would soon be seen to be impossible, 
and the stereometric line of investigation would prove unfruitful 
and be abandoned accordingly. 

The two problems of the duplication of the cube and the 
trisection of an angle, leading in one case to a pure cubic equation 
and in the other to a mixed cubic, are then the only problems 
leading to cubic equations which we can be certain that the Greeks 
had occupied themselves with up to the time of the discovery of the 
conic sections. Menaechmus, who discovered these, showed that 
they could be successfully used for finding the two mean propor- 
tionals and therefore for solving the pure cubic equation, and the 


CXXXVI1Li INTRODUCTION. 


next question is whether it had been proved before the date of 
Aristaeus’ Solid Loci that the trisection of an angle could be 
effected by means of the same conics, either in the form of the 
νεῦσις above described directly and without the reduction to a cubic 
equation, or in the form of the subsidiary cubic (8). Now (1) the 
solution of the cubic would be somewhat difficult in the days when 
conics were still a new thing. The solution of the equation () as 
such would involve the drawing of the conics which we should 
represent by the equations 
xy =a’, 
ba = 3a’ + 3by — y’, 


and the construction would be decidedly more difficult than that 
used by Archimedes in connexion with his cubic, which only requires 
the construction of the conics 


2 


6? 
C= — 
a 


Y) 


(a- x) y=ac; 


hence we can hardly assume that the trisection of an angle in the 
form of the subsidiary cwbic equation was solved by means of conics 
before 320 B.c. (2) The angle may have been trisected by means 
of conics in the sense that the νεῦσις referred to was effected by 
drawing the curves (a), ie. a rectangular hyperbola and a circle. 
This could easily have been done before the date of Aristaeus ; but 
if the assignment of the name ‘solid loci’ to conics had in view their 
applicability to the direct solution of the problem in this manner 
without any reference to the cubic equation, or simply because 
the problem had been before proved to be ‘solid’ by means of the 
reduction to that cubic, then there does not appear to be any 
reason why the Quadratrix, which had been used for the same- 
purpose, should not at the time have been also regarded as a ‘solid 
locus,’ in which case Aristaews could hardly have appropriated the 
latter term, in his work, to conics alone. (3) The only remaining 
alternative consistent with Zeuthen’s view of the origin of the 
name ‘solid locus’ appears to be to suppose that conics were so 
called simply because they gave a means of solving one ‘solid 
problem,’ viz. the doubling of the cube, and not a problem of the 
more general character corresponding to a mixed cubic equation, in 
which case the justification for the general name ‘solid locus’ could 
only be admitted on the assumption that it was adopted at a time 


CUBIC EQUATIONS. CXXX1X 


when the Grecks were still hoping to be able to reduce the general 
cubic equation to the pure form. I think however that the 
traditional explanation of the term is more natural than this 
would be. Conics were the first curves of general interest for 
the description of which recourse to solid figures was necessary as 
distinct from the ordinary construction of plane figures in a plane*; 
hence the use of the term ‘solid locus’ for conics on the mere ground 
of their solid origin would be a natural way of describing the new 
class of curves in the first instance, and the term would be likely 
to remain in use, even when the solid origin was no longer thought 
of, just as the individual conics continued to be called “ sections of 
a right-angled, obtuse-angled, and acute-angled” cone respectively. 
While therefore, as I have said, the two problems mentioned 
might naturally have been called ‘solid problems’ before the dis- 
covery of ‘solid loci,’ I do not think there is sufficient evidence 
to show that ‘solid problem’ was then or later a technical term 
for a problem capable of reduction to a cubic equation in the sense 
of implying that the geometrical equivalent of the general cubic 
equation was investigated for its own sake, independently of its 
applications, and that it ever occupied such a recognised position 
in Greek geometry that a problem would be considered solved so 
soon as it was reduced to a cubic equation. If this had been so, 
and if the technical term for such a cubic was ‘solid problem,’ I 
find it hard to see how Archimedes could have failed to imply some- 
thing of the kind when arriving at his cubic equation. Instead of 
this, his words rather suggest that he had attacked it as res integra. 
Again, if the general cubic had been regarded over any length of 
time as a problem of independent interest which was solved by 
means of the intersections of conics, the fact could hardly have been 
unknown to Nicoteles who is mentioned in the preface to Book Iv. 
of the Conics of Apollonius as having had a controversy with Conon 
respecting the investigations in which the latter discussed the maxi- 
mum number of points of intersection between two conics. Now 
Nicoteles is stated by Apollonius to have maintained that no use 


* Tt is true that Archytas’ solution of the problem of the two mean propor- 
tionals used a curve of double curvature drawn on a cylinder; but this was not 
such a curve as was likely to be investigated for itself or even to be regarded as 
a locus, strictly speaking; hence the solid origin of this isolated curve would 
not be likely to suggest objections to the appropriation of the term ‘solid locus’ 
to conics. 


ex! INTRODUCTION. 


could be made of the discoveries of Conon for διορισμοί; but it seems 
incredible that Nicoteles could have made such a statement, even for 
controversial purposes, if cubic equations then formed a recognised 
class of problems for the discussion of which the intersections of 
conics were necessarily all-important. 

I think therefore that the positive evidence available will not 
justify us in accepting the conclusions of Zeuthen except to the 
following extent. 


1. Pappus’ explanation of the meaning of the term ‘plane 
problem’ (ἐπίπεδον πρόβλημα) as used by the ancients can hardly 
be right. Pappus says, namely, that “problems which can be 
solved by means of the straight line and circle may properly be 
called plane (λέγοιτ᾽ ἂν εἰκότως ἐπίπεδα) ; for the lines by means of 
which such problems are solved have their origin in a plane.” The 
words ‘may properly be called” suggest that, so far as plane 
problems were concerned, Pappus was not giving the ancient 
definition of them, but his own inference as to why they were 
called ‘plane.’ The true significance of the term is no doubt, as 
Zeuthen says, not that straight lines and circles have their origin 
in a plane (which would be equally true of some other curves), but 
that the problems in question admitted of solution by the ordinary 
plane methods of transformation of areas, manipulation of simple 
equations between areas, and in particular the application of areas. 
In other words, plane problems were those which, if expressed 
algebraically, depend on equations of a degree not higher than the 
second. 


2. When further problems were attacked which proved to be 
beyond the scope of the plane methods referred to, it would be 
found that some of such problems, in particular the duplication 
of the cube and the trisection of an angle, were reducible to simple . 
equations between volwmes instead of equations between areas ; and 
it is quite possible that, following the analogy of the distinction 
existing in nature between plane figures and solid figures (an analogy 
which was also followed in the distinction between numbers as ‘plane’ 
and ‘solid’ expressly drawn by Euclid), the Greeks applied the term 
‘solid problem’ to such a problem as they could reduce to an 
equation between volumes, as distinct from a ‘plane problem’ 
reducible to a simple equation between areas. 


3. The first ‘solid problem’ in this sense which they succeeded 


CUBIC EQUATIONS. exli 


in solving was the multiplication of the cube, corresponding to the 
solution of a pure cubic equation in algebra, and it was found that 
this could be effected by means of curves obtained by making plane 
sections of a solid figure, namely the cone. Thus curves having a 
solid origin were found to solve one particular solid problem, which 
could not but seem an appropriate result ; and hence the conic, as 
being the simplest curve so connected with a solid problem, was 
considered to be properly termed a ‘solid locus,’ whether because of 
its application or (more probably) because of its origin. 


4. Further investigation showed that the general cubic equation 
could not be reduced, by means of stereometric methods, to the 
simpler form, the pure cubic; and it was found necessary to try 
the method of conics directly either (1) upon the derivative cubic 
equation or (2) upon the original problem which led to it. In 
practice, as e.g. in the case of the trisection of an angle, it was 
found that the cubic was often more difficult to solve in that 
manner than the original problem was. Hence the reduction of 
it to a cubic was dropped as an unnecessary complication, and 
the geometrical equivalent of a cubic equation stated as an in- 
dependent problem never obtained a permanent footing as the 
‘solid problem’ par excellence. 


5. It followed that solution by conics came to be regarded as 
the criterion for distinguishing a certain class of problem, and, as 
conics had retained their old name of ‘solid loci,’ the corresponding 
term ‘solid problem’ came to be used in the wider sense in which 
Pappus interprets it, according to which it includes a problem 
depending on a biquadratic as well as a problem reducible to a 
cubic equation. 


6. The terms ‘linear problem’ and ‘linear locus’ were then 
invented on the analogy of the other terms to describe respectively 
a problem which could not be solved by means of straight lines, 
circles, or conics, and a curve which could be used for solving such 
a problem, as explained by Pappus. 


CHAPTER VII. 
ANTICIPATIONS BY ARCHIMEDES OF THE INTEGRAL CALCULUS. 


Ir has been often remarked that, though the method of exhaustion 
exemplified in Euclid xu. 2 really brought the Greek geometers face 
to face with the infinitely great and the infinitely small, they 
never allowed themselves to use such conceptions. It is true that 
Antiphon, a sophist who is said to have often had disputes with 
Socrates, had stated* that, if one inscribed any regular polygon, 
say a square, in a circle, then inscribed an octagon by constructing 
isosceles triangles in the four segments, then inscribed isosceles 
triangles in the remaining eight segments, and so on, ‘until the 
whole area of the circle was by this means exhausted, a polygon 
would thus be inscribed whose sides, in consequence of their small- 
ness, would coincide with the circumference of the circle.” But as 
against this Simplicius remarks, and quotes Eudemus to the same 
effect, that the inscribed polygon will never coincide with the 
circumference of the circle, even though it be possible to carry 
the division of the area to infinity, and to suppose that it would 
is to set aside a geometrical principle which lays down that magni- 
tudes are divisible ad infinitum}. The time had, in fact, not come 
for the acceptance of Antiphon’s idea, and, perhaps as the result of - 
the dialectic disputes to which the notion of the infinite gave rise, 
the Greek geometers shrank from the use of such expressions as 
infinitely great and infinitely small and substituted the idea of things 
greater or less than any assigned magnitude. Thus, as Hankel says t, 
they never said that a circle is a polygon with an infinite number of 


* Bretschneider, p. 101. 

+ Bretschneider, p. 102. 

+ Hankel, Zur Geschichte der Mathematik im Alterthum und Mittelalter, 
Ρ. 123. 


ARCHIMEDES ANTICIPATIONS OF THE INTEGRAL CALCULUS, exlili 


infinitely small sides ; they always stood still before the abyss of the 
infinite and never ventured to overstep the bounds of clear con- 
ceptions. They never spoke of an infinitely close approximation or 
a limiting value of the sum of a series extending to an infinite 
number of terms. Yet they must have arrived practically at such 
a conception, e.g., in the case of the proposition that circles are to 
one another as the squares on their diameters, they must have been 
in the first instance led to infer the truth of the proposition by the 
idea that the circle could be regarded as the limit of an inscribed 
regular polygon with an indefinitely increased number of corre- 
spondingly small sides, They did not, however, rest satisfied with 
such an inference ; they strove after an irrefragable proof, and this, 
from the nature of the case, could only be an indirect one. <Ac- 
cordingly we always find, in proofs by the method of exhaustion, 
a demonstration that an impossibility is involved by any other 
assumption than that which the proposition maintains. Moreover 
this stringent verification, by means of a double reductio ad ab- 
surdum, is repeated in every individual instance of the use of the 
method of exhaustion ; there is no attempt to establish, in lieu of 
this part of the proof, any general propositions which could be 
simply quoted in any particular case. 

The above general characteristics of the Greek method of 
exhaustion are equally present in the extensions of the method 
found in Archimedes. To illustrate this, it will be convenient, 
before passing to the cases where he performs genuine integrations, 
to mention his geometrical proof of the property that the area of a 
parabolic segment is four-thirds of the triangle with the same base 
and vertex. Here Archimedes exhausts the parabola by continually 
drawing, in each segment left over, a triangle with the same base 
and vertex as the segment. If A be the area of the triangle so 
inscribed in the original segment, the process gives a series of areas 


Ag An {ΠΕ eas. 
and the area of the segment is really the sum of the infinite series 
A {1444 (D8+ (+. 
But Archimedes does not express it in this way. He first proves 


that, if A,, A,,...4,, be any number of terms of such a series, so that 
A, =4A,, A,=4A,,..., then 


A,+A,+A,+...+4,+44, = $4), 
or Aa GY oe HG) SSG) 4 - τὶ 


exliv INTRODUCTION. 


Having obtained this result, we should nowadays suppose n to 
increase indefinitely and should infer at once that (4)"-? becomes 
indefinitely small, and that the limit of the sum on the left-hand side 
is the area of the parabolic segment, which must therefore be equal 
to 44. Archimedes does not avow that he inferred the result in 
this way; he merely states that the area of the segment is equal 
to 4A, and then verifies it in the orthodox manner by proving that 
it cannot be either greater or less than $A. 

I pass now to the extensions by Archimedes of the method 
of exhaustion which are the immediate subject of this chapter. It 
will be noticed, as an essential feature of all of them, that 
Archimedes takes both an inscribed figure and a circumscribed 
figure in relation to the curve or surface of which he is investigating 
the area or the solid content, and then, as it were, compresses the 
two figures into one so that they coincide with one another and 
with the curvilinear figure to be measured; but again it must 
be understood that he does not describe his method in this way or 
say at any time that the given curve or surface is the limiting form 
of the circumscribed or inscribed figure. I will take the cases 
in the order in which they come in the text of this book. 


1. Surface of a sphere or spherical segment. 


The first step is to prove (On the Sphere and Cylinder 1. 21, 22) 
that, if in a circle or a segment of a circle there be inscribed 
polygons, whose sides AB, BC, CD, ... are all equal, as shown 
in the respective figures, then 


(a) for the circle 
(BB τοῦδέ +..:) AA HAR 1.83. 
(Ὁ) for the segment 
(BB'+CC'+...+ KK'+ IM): AM=A'B: BA. 


Next it is proved that, if the polygons revolve about the 
diameter AA’, the surface described by the equal sides of the 
polygon in a complete revolution is [1. 24, 35] 


(a) equal to a circle with radius Ὄ AB(BB’+CC gray YY) 


or (ὦ) equal to a circle with radius JAB (BB +CC' +, ASL 


Therefore, by means of the above proportions, the surfaces 
described by the equal sides are seen to be equal to 


ARCHIMEDES’ ANTICIPATIONS OF THE INTEGRAL CALCULUS. exlv 


(a) a circle with radius AAR AB, 

and (Ὁ) a circle with radius ,/ AM. 4΄8; 

they are therefore respectively [1. 25, 37] less than 
(a) acircle with radius 4 Μ΄, 
(6) a circle with radius AL. 


Archimedes now proceeds to take polygons circumscribed to the 
circle or segment of a circle (supposed in this case to be less than a 
semicircle) so that their sides are parallel to those of the inscribed 
polygons before mentioned (cf. the figures on pp. 38, 51); and he 
proves by like steps [1. 30, 40] that, if the polygons revolve about the 
diameter as before, the surfaces described by the equal sides during 
a complete revolution are greater than the same circles respectively. 

Lastly, having proved these results for the inscribed and 
circumscribed figures respectively, Archimedes concludes and proves 
[1. 33, 42, 43] that the surface of the sphere or the segment of the 
sphere is equal to the first or the second of the circles respectively. 

In order to see the effect of the successive steps, let us express 
the several results by means of trigonometry. If, in the figures on 
pp. 33, 47 respectively, we suppose 4n to be the number of sides in 
the polygon inscribed in the circle and 2n the number of the equal 
sides in the polygon inscribed in the segment, while in the latter 
case the angle AOL is denoted by a, the proportions given above 
are respectively equivalent to the formulae * 


9 


= T - ΧΙ - T T 
sin =— + sin — +...+sin(2n—1) — =cot-, 
Qn Qn Qn 4n 
α΄ ἢ τ ὭΧΩ : a : 
2 4sin—+sin —+...+sin(n—1)—}+sina 
vi 1 γὺ α 
and ΞΟ 
l—cosa Qn 


Thus the two proportions give in fact a summation of the series 
sin 6+ sin 26+ ...+sin (n—1)6 


both generally where 70 is equal to any angle a less than z, and in 
the particular case where 7 is even and 6=7/n. 

Again, the areas of the circles which are equal to the surfaces 
described by the revolution of the equal sides of the inscribed 


* These formulae are taken, with a slight modification, from Loria, Il periodo 
aureo della geometria greca, p. 108. 


Ἐΐ A. k 


exlvi INTRODUCTION. 


polygons are respectively (if ὦ be the radius of the great circle 
of the sphere) 


9 

5 T Ξ π a ate - T π 

47a? sin - Ἐπ =—+sin5— +... 51π0ὸ ἰδ or Ἵππον 
4n 2n 2n 2n 4n 


“: 


and 


9) 
. α τος ἰοῦ . «α . α . 
wa’. 2 sin =| 2 jsin—+sin — +... + sin(n—1)—} +sina |, 
2n n n n 


a 
or πα". 2 cos ae (1 — cos a). 
In 


The areas of the circles which are equal to the surfaces described 
by the equal sides of the circwmscribed polygons are obtained from 
the areas of the circles just given by dividing them by cos’ z/4n and 
cos’ a/2n respectively. 

Thus the results obtained by Archimedes are the same as would 
be obtained by taking the limiting value of the above trigonometri- 
cal expressions when 7 is indefinitely increased, and when therefore 
cos 7/4n and cos αἰ 27. are both unity. 

But the first expressions for the areas of the circles are (when 7 
is indefinitely increased) exactly what we represent by the 
integrals 


Tr 
47a”. 4 i sin 6d6, or 47a’, 
0 
9 vd 2 9 
and πα΄. | 2 sin @d6, or 27a’ (1 -- 008 a). 
0 


Thus Archimedes’ procedure is the equivalent of a genuine 
integration in each case. 

2. Volume of a sphere or a sector of a sphere. 

The method does not need to be separately set out in detail here, 
because it depends directly on the preceding case: The investiga- 
tion proceeds concurrently with that of the surface of a sphere or a 
segment of a sphere. The same inscribed and circumscribed figures 
are used, the sector of a sphere being of course compared with the 
solid figure made up of the figure inscribed or circumscribed to the 
segment and of the cone which has the same base as that figure and 
has its vertex at the centre of the sphere. It is then proved, 
(1) for the figure inscribed or circumscribed to the sphere, that its 
volume is equal to that of a cone with base equal to the surface of 
the figure and height equal to the perpendicular from the centre of 
the sphere on any one of the equal sides of the revolving polygon, 
(2) for the figure inscribed or circumscribed to the sector, that the 


ARCHIMEDES ANTICIPATIONS OF THE INTEGRAL CALCULUS. exlvii 


volume is equal to that of a cone with base equal to the surface of 
the portion of the figure which is inscribed or circumscribed to the 
segment of the sphere included in the sector and whose height is the 
perpendicular from the centre on one of the equal sides of the 
polygon. 

Thus, when the inscribed and circumscribed figures are, so 
to speak, compressed into one, the taking of the limit is practically 
the same thing in this case as in the case of the surfaces, the 
resulting volumes being simply the before-mentioned surfaces 
multiplied in each case by 4a. 


3. Area of an ellipse. 


This case again is not strictly in point here, because it does 
not exhibit any of the peculiarities of Archimedes’ extensions of 
the method of exhaustion. That method is, in fact, applied in 
the same manner, mutatis mutandis, as in Eucl. xu. 2. There 
is no simultaneous use of inscribed and circumscribed figures, but 
only the simple exhaustion of the ellipse and auxiliary circle by 
increasing to any desired extent the number of sides in polygons 
inscribed to each (On Conoids and Spheroids, Prop. 4). 


4. Volume of a segment of a paraboloid of revolution. 


Archimedes first states, as a Lemma, a result proved incidentally 
in a proposition of another treatise (On Spirals, Prop. 11), viz. that, 
if there be » terms of an arithmetical progression h, 2h, 3h, ..., then 


h+2h+ 3h+...+nh>tnh 
and Pe SE aa 


Next he inscribes and circumscribes to the segment of the 
paraboloid figures made up of small cylinders (as shown in the figure 
of On Conoids and Spheroids, Props. 21, 22) whose axes lie along 
the axis of the segment and divide it into any number of equal 
parts. If ὁ is the length of the axis 4D of the segment, and if 
there are 7 cylinders in the circumscribed figure and their axes are 
each of length h, so that e=nh, Archimedes proves that 


1 cylinder CE _ wh 
(1) inscribed fig. h+2h+3h+...+(n-1p)h 
> 2, by the Lemma, 
cylinder CH nh 


2 ee nen 
eae (2) circumscribed fig, h+2h+3h+...+nh 


«Ὁ. 


k2 


exlvill INTRODUCTION. 


Meantime it has been proved [Props. 19, 20] that, by increasing 
m sufficiently, the inscribed and circumscribed figure can be made 
to differ by less than any assignable volume. It is accordingly 
concluded and proved by the usual rigorous method that 


(cylinder CZ) = 2 (segment), 
so that (segment ABC) = 3 (cone ABC). 


The proof is therefore equivalent to the assertion, that if h is 
indefinitely diminished and m indefinitely increased, while nh remains 
equal to ὁ, 

limit of h{h + 2h+3h+...+(n—-1)h} =he'; 


Cc 
i ada = 1c. 
᾿ 2 


Thus the method is essentially the same as ours when we 
express the volume of the segment of the paraboloid in the form 


that is, in our notation, 


eo 
K | ydx, 
Jo 


where « is a constant, which does not appear in Archimedes’ result 
for the reason that he does not give the actual content of the 
segment of the paraboloid but only the ratio which it bears to the 
circumscribed cylinder. 


5. Volume of a segment of a hyperboloid of revolution. 


The first step in this case is to prove [On Conoids and Spheroids, 
Prop. 2] that, if there be a series of m terms, 


ah+h?, a.2h+(2h), a.3h+ (3h), ... a.nh+ (nhy’, 


and if (ah+h?)+{'a.2h+(2h)+...4+fa.nh+(nh)} =S8,, 


then fa. wh + (nl), <(a+nh)| (+S) 
ee (8). 
nh 
ane nia. nh + (nh)?}/Sy— -17 (a te mt) | (5+ 2) 


Next [Props. 25, 26] Archimedes draws inscribed and cireum- 
scribed figures made up of cylinders as before (figure on p. 137), and 


ARCHIMEDES’ ANTICIPATIONS OF THE INTEGRAL CALCULUS. cxlix 
proves that, if AD is divided into m equal parts of length ἡ, so that 
nh = AD, and if AA’ =a, then 


cylinder HB’ πίω. nh + (nh)? 
inscribed figure | Spon 


> (a+ nh) | (5 + =) : 


cylinder Β΄ —s n{a. nh + (nh)’} 
circumscribed fig. | δ 


a nh 
<(a+nh)/(=-+—)- 
(a+nhy/(5 +5) 
The conclusion, arrived at in the same manner as before, is that 


cylinder LB’ Ψ a nh 
segment ABB’ — (Ὁ un) (5 3 ) ; 


and 


This is the same as saying that, if nh =b, and if h be indefinitely 
diminished while x is indefinitely increased, 


limit of n (ab +.6°)/S,=(a+8) / ( + 3) 
eee b tie a 
or limit of 5 ἊΣ ΞΡ (ξ -Ὲ 5)" 
Now 8S,=a(h+2h+...+nh) + {h?+ (2h)? +...+(nhy}, 
sothat AS,=ah(h+2h+...+nh)+h th? + (2h)? +... + (nh). 
The limit of the last expression is what we should write as 
b 
i (ax + x) da, 
0 
which is equal to ὑ" (5 + >) : 


and Archimedes has given the equivalent of this integration. 


6. Volume of a segment of a spheroid. 
Archimedes does not here give the equivalent of the integration 
(ax — x”), 
0 


presumably because, with his method, it would have required yet 
another lemma corresponding to that in which the results (@) above 
are established. 


el INTRODUCTION. 


Suppose that, in the case of a segment less than half the spheroid 
(figure on p. 142), AA’=a, CD=4c, AD=b; and let AD be divided 
into n equal parts of length h. 

The gnomons mentioned in Props. 29, 30 are then the differences 
between the rectangle cb + 6” and the successive rectangles 

ch+h?, ¢.2h+(2h)’, ... e.(n—1)h+{(n—- 1) Ay, 
and in this case we have the conclusions that (if S,, be the sum of 
n terms of the series representing the latter rectangles) 
cylinder HB’ π(οδ- δὴ) 
inscribed figure ἡ (cb +b) — S, 


76 2b 
τ χὰ Goa) 


_cylinder EB' n (cb + 6°) 
circumscribed fig. ἡ (cb + 6°) — δ᾽... 


Bt 


: sane cylinder LB’ _ 
and in the limit eon Gia (e+ »| (5+ 


and 


Accordingly we have the limit taken of the expression 
n (οὗ + δ) —S, την, δ 
πὶ (οὗ -- 6) ᾿ γι (οὗ + 65) 


and the integration performed is the same as that in the case of the 
hyperboloid above, with ὁ substituted for a, 


Archimedes discusses, as a separate case, the volume of half. a 
spheroid [Props. 27, 28]. It differs from that just given in that ὁ 
vanishes and b= 4a, so that it is necessary to find the limit of 
h? + (2h)? + (3h)? +... + (nh)? | 
n (nh)? et 


and this is done by means of a corollary to the lemma given on 
pp. 107—9 [On Spirals, Prop. 10] which proves that 


h? + (2h)? +... + (nh)? > En (nh)?, 
and h? + (2h)? + "τὸ 1) h}? «1 (nh)’. 


The limit of course vor. to the integral 


b 
i a’dx = 46°. 
0 


ARCHIMEDES’ ANTICIPATIONS OF THE INTEGRAL CALCULUS. cli 


7. Area of a spiral. 


(1) Archimedes finds the area bounded by the first complete 
turn of a spiral and the initial line by means of the proposition just 
quoted, viz. 

Δ + (2h)? +... + (nh) > kn (nh), 
h? + (2h)? +... + {(n—1) hi? < dn (nh)?. 


He proves [Props. 21, 22, 23] that a figure consisting of similar 
sectors of circles can be circumscribed about any arc of a spiral such 
that the area of the circumscribed figure exceeds that of the spiral 
by less than any assigned area, and also that a figure of the same 
kind can be inscribed such that the area of the spiral exceeds that 
of the inscribed figure by less than any assigned area. Then, lastly, 
he circumscribes and inscribes figures of this kind [ Prop. 24]; thus 
e.g. in the circumscribed figure, if there are ~ similar sectors, the 
radii will be m lines forming an arithmetical progression, as h, 2h, 
3h, ... nh, and nh will be equal to a, where a is the length inter- 
cepted on the initial line by the spiral at the end of the first turn. 
Since, then, similar sectors are to one another as the square of their 
radii, and 7 times the sector of radius nh or a is equal to the circle 
with the same radius, the first of the above formulae proves that 


(circumscribed fig.) > 47a’. 


A similar procedure for the inscribed figure leads, by the use of the 
second formula, to the result that 


(inscribed fig.) < 47a’. 
The conclusion, arrived at in the usual manner, is that 
(area of spiral) = 47a’ ; 


and the proof is equivalent to taking the limit of 
T 9 9 
A [5 + (2h)? +... + τοὺ — 1) A}? ] 
f re 21. f 12 
or 0 - Lh + (2h +... + {(n— 1) A}? ], 
which last limit we should express as 


a 
T 
—| 2°dx=t7a" 
(10 


eli INTRODUCTION. 


{It is clear that this method of proof equally gives the area 
bounded by the spiral and any radius vector of length 6 not being 
greater than a; for we have only to substitute wb/a for z, and to 
remember that in this case nh=b. We thus obtain for the area 


T 


b 
-- | ada, or 47b*/a. | 
ao 


(2) To find the area bounded by an are on any turn of the 
spiral (not being greater than a complete turn) and the radii 
vectores to its extremities, of lengths ὦ and ὁ say, where 6: ὦ, 
Archimedes uses the proposition that, if there be an arithmetic 
progression consisting of the terms 


b, b+h, b+ 2h, ... δ- (ἡ -- 1), 


and if δ΄, Ξε 0+ (b+h)? + (b+ 2h)? +... 4+ {b+ (ἡ -- 1)λ}, 


(n—1){b+(m—1)h}? ἐδ + (m— 1) hj? 
nee S,-8 ει ὦ 
—1){b+(n-1)h'? {b+ (n—1)h}? 
d (ays es al νἀ ππ" 
τὰ ἘΣ * + (m—1)yhib+ki(n—1) hp 


[On Spirals, Prop. 11 and note. ] 

Then in Prop. 26 he circumscribes and inscribes figures consisting 
of similar sectors of circles, as before. There are n—1 sectors in 
each figure and therefore 7 radii altogether, including both ὦ and ὁ, 
so that we can take them to be the terms of the arithmetic progres- 
sion given above, where {b+(u—1)h}=c. It is thus proved, by 
means of the above inequalities, that 


__ sector OB'C ¥ {b+ (n—1)h}? _ sector OBC 
circumscribed fig. — {b+(n—1)h}b+44(n—1)h\? inser. fig.” 


and it is concluded after the usual manner that 


sector OB'C {b+ (n—1)h? 


spiral OBU ~ {b+(n—1)htb +4 f(n—1) hi? 


9 


— σ' 
δ -Ἐ1 (6-- δ)" 


Remembering that ἡ -- 1 -- (ὁ -- ὁὴ),}, we see that the result is the 


ARCHIMEDES’ ANTICIPATIONS OF THE INTEGRAL CALCULUS. ΟΠ 


same thing as proving that, in the limit, when x becomes indefinitely 
great and ἢ indefinitely small, while ὦ + (n—1)h=c, 


limit of A[b°+(b+A)? +... + {b+ (m— 2)}}}} 
Ξε (ὁ -- ὁ) {ch +4 (e—b)"} 


Ξε ἡ (οὗ -- δ); 
that is, with our notation, 


(3) Archimedes works out separately [Prop. 25], by exactly 
the same method, the particular case where the area is that described 
in any one complete turn of the spiral beginning from the initial 
line. This is equivalent to substituting (x—1)a for ὦ and na for ¢, 
where ὦ is the radius vector to the end of the first complete turn of 
the spiral. 

It will be observed that Archimedes does not use the result 


corresponding to 
6 6 b 
i wda— | eda = | x da. 
0 b 0 


8. Area of a parabolic segment. 


Of the two solutions which Archimedes gives of the problem of 
squaring a parabolic segment, it is the mechanical solution which 
gives the equivalent of a genuine integration. In Props. 14, 15 of 
the Quadrature of the Parabola it is proved that, of two figures 
inscribed and circumscribed to the segment and consisting in each 
case of trapezia whose parallel sides are diameters of the parabola, 
the inscribed figure is less, and the circumscribed figure greater, 
than one-third of a certain triangle (ZqQ in the figure on p. 242). 
Then in Prop. 16 we have the usual process which is equivalent to 
taking the limit when the trapezia become infinite in number and 
their breadth infinitely small, and it is proved that 


(area of segment) =} A £qQ. 


The result is the equivalent of using the equation of the parabola 
referred to Qg as axis of « and the diameter through Q as axis of 
ae VIZ: 

py =x (2a — 2), 
which can, as shown on p. 236, be obtained from Prop. 4, and finding 


2a 


yda, 
0 


cliv INTRODUCTION. 


where y has the value in terms of # given by the equation ; and of 


course 
1 [2α 
=f (2ax —x*) dx = 
P Jo 
The equivalence of the method to an integration can also be 
seen thus. It is proved in Prop. 16 (see figure on p. 244) that, if 
gE be divided into 2 equal parts and the construction of the 
proposition be made, QYq is divided at O,, O,,... into the same 
number of equal parts. The area of the circumscribed figure is then 
easily seen to be the sum of the areas of the triangles 


QqF, QR,F,, QR,F,, tee 


403 
3p ; 


that is, of the areas of the triangles 
QqF, QO,R,, 0. ;, ... 


Suppose now that the area of the triangle qf is denoted by A, and 
it follows that 


- 7): ΞΕ ΟἿΣ 
(circumscribed fig.) = Δ {1 + @ it) + ine ) eee = 


» 9 


nm ne 7.0 


Similarly we obtain 


. . 1 9AQ 
(inscribed fig.) = PORE A fA? + 27A? +... + (n— 1) At. 


Taking the limit we have, if A denote the area of the triangle ZqQ, 
so that A=nA, 


1 4 
(area of segment) = Ἵ: | A’dA 
Jo 


= 


If the conclusion be regarded in this manner, the integration is 
the same as that which corresponds to Archimedes’ squaring of the 
spiral. 


CHAPTER VIII. 
THE TERMINOLOGY OF ARCHIMEDES. 


So far as the language of Archimedes is that of Greek geometry 
in general, it must necessarily have much in common with that of 
Euclid and Apollonius, and it is therefore inevitable that the 
present chapter should repeat many of the explanations of terms of 
general application which I have already given in the corresponding 
chapter of my edition of Apollonius’ Conics*. But I think it will 
be best to make this chapter so far as possible complete and self- 
contained, even at the cost of some slight repetition, which will 
however be relieved (1) by the fact that all the particular phrases 
quoted by way of illustration will be taken from the text of 
Archimedes instead of Apollonius, and (2) by the addition of a large 
amount of entirely different matter corresponding to the great 
variety of subjects dealt with by Archimedes as compared with the 
limitation of the work of Apollonius to the one subject of conics. 

One element of difficulty in the present case arises out of the 
circumstance that, whereas Archimedes wrote in the Doric dialect, 
the original language has been in some books completely, and in others 
partially, transformed into the ordinary dialect of Greek. Uni- 
formity of dialect cannot therefore be preserved in the quotations 
about to be made; but I have thought it best, when explaining 
single words, to use the ordinary form, and, when illustrating their 
use by quoting phrases or sentences, to give the latter as they appear 
in Heiberg’s text, whether in Doric or Attic in the particular case. 
Lest the casual reader should imagine the paroxytone words εὐθείαι, 
διαμέτροι, πεσείται, πεσούνται, ἐσσείται, δυνάνται, ἁπτέται, καλείσθαι, 
κείσθαι and the like to be misprints, I add that the quotations in 
Doric from Heiberg’s text have the unfamiliar Doric accents. 

Τ shall again follow the plan of grouping the various technical 


* Apollonius of Perga, pp. clvii—clxx. 


elvi INTRODUCTION. 


terms under certain general headings, which will enable the Greek 
term corresponding to each expression in the ordinary mathematical 
phraseology of the present day to be readily traced wherever such 
a Greek equivalent exists. 


Points and lines. 


A point is σημεῖον, the point B τὸ Β σημεῖον or τὸ B simply; a 
point on (a line or curve) σημεῖον ἐπί (with gen.) or ἐν; ὦ point 
raised above (a plane) σημεῖον μετέωρον ; any two points whatever 
being taken δύο σημείων λαμβανομένων ὁποιωνοῦν. 

At a point (e.g. of an angle) πρός (with dat.), having its vertex at 
the centre of the sphere κορυφὴν ἔχων πρὸς TO κέντρῳ τῆς σφαίρας ; of 
lines meeting in a point, touching or dividing αὐ a point, etc., κατά 
(with acc.), thus AE is bisected at Zis a AE δίχα τεμνέται κατὰ τὸ Z; 
of a point falling on or being placed on another ἐπί or κατά (with 
acc.), thus Z will fall on T, τὸ μὲν Z ἐπὶ τὸ Τ' πεσείται, so that ἘΣ lies 
on Δ, ὥστε τὸ μὲν E κατὰ τὸ A κείσθαι. 

Particular points are eatremity πέρας, vertex κορυφή, centre 
κέντρον, point of division διαίρεσις, point of meeting σύμπτωσις, pornt 
of section τομή, point of bisection διχοτομία, the middle point τὸ 
μέσον ; the points of division H, 1, K, ta τῶν διαιρεσίων σαμεῖα τὰ H, 
I, K; let B be its middle point μέσον δὲ αὐτᾶς ἔστω τὸ B; the point of 
section in which (a circle) cuts a toa, καθ᾽ ἃν τέμνει. 

A line is γραμμή, a curved line καμπύλη γραμμή, a straight line 
εὐθεῖα with or without γραμμή. The straight line @IKA, a OIKA 
εὐθεῖα ; but sometimes the older expression is used, the straight line 
on which (ἐπί with gen. or dat. of the pronoun) are placed certain 
letters, thus let it be the straight line M, ἔστω ἐφ᾽ ἃ τὸ M, other 
straight lines K, A, ἄλλαι γραμμαί, ἐφ᾽ av ta K, A. The straight 
lines between the points at μεταξὺ τῶν σημείων εὐθεῖαι, of the lines 
which have the same extremities the straight line is the least τῶν τὰ 
αὐτὰ πέρατα ἐχουσῶν γραμμῶν ἐλαχίστην εἶναι τὴν εὐθεῖαν, straight lines 
cutting one another εὐθείαι τεμνούσαι ἀλλάλας. 

For points in relation to lines we have such expressions as the 
following: the points T, ©, M are on a straight line ἐπ᾽ εὐθείας ἐστὶ 
τὰ Τ', @, M σαμεῖα, the point of bisection of the straight line containing 
the centres of the middle magnitudes ἃ διχοτομία τᾶς εὐθείας τᾶς 
ἐχούσας τὰ κέντρα τῶν μέσων μεγεθέων. A very characteristic phrase 
for αὐ a point which divides the straight line in such a proportion 
that... is ἐπὶ τᾶς εὐθείας διαιρεθείσας wore...; similarly ἐπὶ τᾶς XE 


THE TERMINOLOGY OF ARCHIMEDES. elvil 


τμαθείσας οὕτως, ὦστε. A certain point will be on the straight line... 
dividing it so that... ἐσσείται ἐπὶ ras εὐθείας...διαιρέον οὕτως τὰν 
εἰρημέναν εὐθεῖαν, Wore.... 

The middle point of a line is often elegantly denoted by an 
adjective in agreement; thus at the middle point of the segment ἐπὶ 
μέσου τοῦ τμάματος, (ὦ line) drawn from T to the middle pot of 
EB, ἀπὸ τοῦ T ἐπὶ μέσαν τὰν EB ἀχθεῖσα, drawn to the middle point of 
the base ἐπὶ μέσαν τὰν βάσιν ἀγομένα. 

A straight line produced is the (straight line) in the same straight 
line with it ἡ ἐπ᾽ εὐθείας αὐτῇ. In the same straight line with the 
avis ἐπὶ tas αὐτᾶς εὐθείας τῷ ἄξονι. Of a straight line falling on 
another line κατά (with gen.) is used, e.g. πίπτουσι κατ᾽ αὐτῆς ; ἐπί 
(with acc.) is also used of a straight line placed on another, thus ¢f 
EH be placed on BA, τεθείσας tas EH ἐπὶ τὰν BA. 


For lines passing through points we find the following ex- 
pressions: will pass through N, ἥξει διὰ τοῦ N ; will pass through the 
centre διὰ τοῦ κέντρου πορεύσεται, will fall through @ πεσείται διὰ Tod 
Θ, verging towards B νεύουσα ἐπὶ τὸ B, pass through the same point 
ἐπὶ τὸ αὐτὸ σαμεῖον ἐρχόνται ; the diagonals of the parallelogram fall 
(i.e. meet) at Θ, κατὰ δὲ τὸ Θ at διαμέτροι τοῦ παραλληλογράμμου 
πίπτοντι ; EZ (passes) through the points bisecting AB, TA, ἐπὶ δὲ τὰν 
διχοτομίαν τἂν AB, TA ἁ EZ. The verb εἰμί is also used of passing 
through, thus ἐσσείται δὴ αὐτὰ διὰ τοῦ Θ. 

For lines in relation to other lines we have perpendicular to 
κάθετος ἐπί (with acc.), parallel to παράλληλος with dat. or παρά 
(with acc.) ; leé KA be (drawn) from K parallel to TA, ἀπὸ τοῦ Καὶ 
παρὰ τὰν TA ἔστω a KA, 


Lines meeting one another συμπίπτουσαι ἀλλήλαις ; the point in 
which ZH, MN produced meet one another and AV, τὸ σημεῖον, καθ᾽ ὃ 
συμβάλλουσιν ἐκβαλλόμεναι ai ZH, MN ἀλλήλαις τε καὶ τῇ AT; so as 
to meet the tangent wore ἐμπεσεῖν τᾷ ἐπιψαυούσᾳ, let straight lines be 
drawn parallel to AT to meet the section of the cone ἄχθων εὐθείαι 
παρὰ τὰν AT ἔστε ποτὶ τὰν τοῦ κώνου τομάν, to draw a straight line to 
meet its circwmference ποτὶ τὰν περιφέρειαν αὐτοῦ ποτιβαλεῖν εὐθεῖαν, 
the line drawn to meet ἃ ποτιπεσοῦσα, let AE, AA be drawn from the 
point A to meet the spiral and produced to meet the circumference of 
the circle ποτιπιπτόντων ἀπὸ τοῦ A σαμείου ποτὶ τὰν ἕλικα ai AE, AA 
καὶ ἐκπιπτόντων ποτὶ τὰν τοῦ κύκλου περιφέρειαν ; wntil it meets OA in 
O, ἔστε κα συμπέσῃ τᾷ ΘΑ κατὰ τὸ O (οὗ a circle). 


elvill INTRODUCTION. 


(The straight line) wili fall outside (i.e. will extend beyond) P, 
ἐκτὸς τοῦ P πεσεῖται ; will fall within the section of the figure ἐντὸς 
πεσούνται TAS τοῦ σχήματος TOMAS. 

The (perpendicular) distance between (two parallel lines) AZ, BH, 
τὸ διάστημα τᾶν AZ, BH. Other ways of expressing distances are the 
following: the magnitudes equidistant from the middle one τὰ ἴσον 
ἀπέχοντα ἀπὸ τοῦ μέσου μεγέθεα, are at equal distances from one 
another ἴσα am ἀλλάλων διέστακεν ; the segments (lengths) on AH 
equal to N, τὰ ἐν τᾷ AH τμάματα ἰσομεγέθεα τᾷ N; greater by one 
segment ἑνὶ τμάματι μείζων. 

The word εὐθεῖα itself is also often used in the sense of distance ; 
cf. the terms πρώτη εὐθεῖα etc. in the book On Spirals, also ἃ εὐθεῖα 
ἅ μεταξὺ τοῦ κέντρου τοῦ ἁλίου Kal τοῦ κέντρου τᾶς yas the distance 
between the centre of the sun and the centre of the earth. 

The word for join is ἐπιζευγνύω or ἐπιζεύγνυμι ; the straight line 
joining the points of contact ἃ τὰς ἁφὰς ἐπιζευγνύουσα εὐθεῖα, BA when 
joined & BA ἐπιζευχθεῖσα ; let EZ join the points of bisection of AA, 
BI, a δὲ EZ ἐπιζευγνυέτω τὰς διχοτομίας τάν AA, BI. In one case 
the word seems to be used in the sense of drawing simply, εἴ κα 


εὐθεῖα ἐπιζευχθῇ ypappa ἐν ἐπιπέδῳ. 


Angles. 


An angle is γωνία, the three kinds of angles are right ὀρθή, acute 
ὀξεῖα, obtuse ἀμβλεῖα ; right-angled etc. ὀρθογώνιος, ὀξυγώνιος, ἀμβλυ- 
γώνιος ; equiangular ἰσογώνιος ; with an even number of angles 
ἀρτιόγωνος Or ἀρτιογώνιος. 

At right angles to ὀρθὸς πρός (with acc.) or πρὸς ὀρθάς (with dat. 
following); thus if a line be erected at right angles to the plane ypappas 
ἀνεστακούσας ὀρθᾶς ποτὶ τὸ ἐπίπεδον, the planes are at right angles to 
one another ὀρθὰ wot ἀλλαλά ἐντι τὰ ἐπίπεδα, being at right angles 
to ABI, zpos ὀρθὰς ὧν τῷ ABI; KT, ZA are at right angles to one 
another ποτ᾽ ὀρθάς ἐντι ἀλλάλαις αἱ KT, BA, to cut at right angles 
τέμνειν πρὸς ὀρθάς. The expression making right angles with is also 
used, e.g. ὀρθὰς ποιοῦσα γωνίας ποτὶ τὰν ΑΒ. 

The complete expression for the angle contained by the lines AH, 
AT is ἅ γωνία & περιεχομένα ὑπὸ τἂν AH, AT’; but there are a great 
variety of shorter expressions, ywvia itself being often understood ; 
thus the angles A, E, A, B, αἱ A, E, A, Β γωνίαι; the angle at Θ, ἃ ποτὶ 
τῷ Θ; the angle contained by AA, AZ, ἃ γωνία ἁ ὑπὸ τᾶν AA, AZ; the 
angle AHT, ἡ ὑπὸ τῶν AHL γωνία, ἡ ὑπὸ AHT (with or without γωνία). 


THE TERMINOLOGY OF ARCHIMEDES. clix 


Making the angle K equal to the angle ©, γωνίαν ποιοῦσα τὰν Καὶ 
ἴσαν τᾷ @; the angle into which the sun fits and which has its vertex 
at the eye γωνία, εἰς av ὃ ἅλιος ἐναρμόζει τὰν κορυφὰν ἔχουσαν ποτὶ τᾷ 
ὄψει; of the sides subtending the right angle (hypotenuses) τᾶν ὑπὸ 
τὰν ὀρθὰν γωνίαν ὑποτεινουσᾶν, they subtend the same angle ἐντὶ ὑπὸ 
τὰν αὐτὰν γωνίαν. 

If a line through an angular point of a polygon divides it 
exactly symmetrically, the opposite angles of the polygon, at ἀπεναντίον 
γωνίαι τοῦ πολυγώνου, are those answering to each other on each side 
of the bisecting line. 


Planes and plane figures. 


A plane ἐπίπεδον; the plane through BA, τὸ ἐπίπεδον τὸ κατὰ 
τὴν BA, or τὸ διὰ τῆς BA, plane of the base ἐπίπεδον τῆς βάσεως, plane 
(ie. base) of the cylinder ἐπίπεδον τοῦ κυλίνδρου : cutting plane ἐπί. 
medov τέμνον, tangent plune ἐπίπεδον ἐπιψαῦον ; the intersection of 
planes is their common section κοινὴ Tomy. 

In the same plane as the circle ἐν τῷ αὐτῷ ἐπιπέδῳ τῷ κύκλῳ. 

Let a plane be erected on UZ at right angles to the plane in which 
AB, TA are ἀπὸ tas ΠΖ ἐπίπεδον ἀνεστακέτω ὀρθὸν ποτὶ τὸ ἐπίπεδον τό, 
ἐν ᾧ ἐντι αἱ ΑΒ, TA. 

The plane surface ἡ ἐπίπεδος (ἐπιφάνεια), a plane segment ἐπίπεδον 
τμῆμα, a plane figure σχῆμα ἐπίπεδον. 

A rectilineal figure εὐθύγραμμον (σχῆμα), a side πλευρά, perimeter 
n περίμετρος, similar ὅμοιος, similarly sitwated ὁμοίως κείμενος. 

To coincide with (when one figure is applied to another), 
ἐφαρμόζειν followed by the dative or ἐπί (with acc.); one part 
coincides with the other ἐφαρμόζει τὸ ἕτερον μέρος ἐπὶ τὸ ἕτερον ; the 
plane through NZ coincides with the plane through AT’, τὸ ἐπίπεδον τὸ 
κατὰ τὰν NZ ἐφαρμόζει τῷ ἐπιπέδῳ τῷ κατὰ τὰν Α΄. The passive is 
also used ; if equal and similar plane figures coincide with one another 


na FF WO ΄, / > fd 3 / ΓΝ ΔῈ 
τῶν ἴσων καὶ ὁμοίων σχημάτων ἐπιπέδων ἐφαρμοζομένων ἐπ᾽ ἀἄλλαλα. 


Triangles. 


A triangle is τρίγωνον, the triangles bounded by (their three 
sides) τὰ περιεχόμενα τρίγωνα ὑπὸ τῶν.... A right-angled triangle 
τρίγωνον ὀρθογώνιον, one of the sides about the right angle μία τῶν περὶ 
τὴν ὀρθήν. The triangle through the aais (of a cone) τὸ διὰ τοῦ ἄξονος 
τρίγωνον. 


elx INTRODUCTION. 


Quadrilaterals. 

A quadrilateral is a fowr-sided figure (τετράπλευρον) as dis- 
tinguished from a fowr-angled figure, τετράγωνον, which means a 
square. A trapezium, τραπέζιον, is in one place more precisely 
described as a trapeziwm having its two sides parallel τραπέζιον τὰς 
δύο πλευρὰς ἔχον παραλλάλους ἀλλάλαις. 

A parallelogram παραλληλόγραμμον ; for a parallelogram on a 
straight line as base ἐπί (with gen.) is used, thus the parallelograms 
on them are of equal height ἐστὶν ἰσούψη τὰ παραλληλόγραμμα τὰ ἐπ᾽ 
αὐτῶν. A diagonal of a parallelogram is διάμετρος, the opposite sides 
of the parallelogram ai κατ᾽ ἐναντίον ToD παραλληλογράμμου πλευραί. 


Rectangles. 


The word generally used for a rectangle is χωρίον (space or area) 
without any further description. As in the case of angles, the 
rectangles contained by straight lines are generally expressed more 
shortly than by the phrase τὰ περιεχόμενα χωρία ὑπό ; either χωρίον 
may be omitted or both χωρίον and περιεχόμενον, thus the rectangle 
AT, TE may be any of the following, τὸ ὑπὸ τῶν AT, TE, τὸ ὑπὸ 
AT, TE, τὸ ὑπὸ ATE, and the rectangle under ΘΚ, AH is τὸ ὑπὸ τῆς 
ΘΚ καὶ τῆς AH. Rectangles Θ, 1, K, A, χωρία ἐν οἷς τὰ (or ἐφ᾽ ὧν 
ἕκαστον τῶν) Θ,1, K, A. 

To apply a rectangle to a straight line (in the technical sense) is 
παραβάλλειν, and παραπίπτω is generally used in place of the passive; 
the participle παρακείμενος is also used in the sense of applied to. In 
each case applying fo a straight line is expressed by παρά (with acc.). 
Examples are, areas which we can apply to a given straight line (i.e. 
which we can transform into a rectangle of the same area) χωρία, ἃ 
δυνάμεθα παρὰ τὰν δοθεῖσαν εὐθεῖαν παραβαλεῖν, let ὦ rectangle be 
applied to each of them παραπεπτωκέτω παρ᾽ ἑκάσταν αὐτᾶν χωρίον : 
if there be applied to each of them a rectangle exceeding by a square 
figure, and the sides of the excesses exceed each other by an equal 
amount (1.6. form an arithmetical progression) εἴ κα παρ᾽ ἑκάσταν 
αὐτᾶν παραπέσῃ τι χωρίον ὑπερβάλλον εἴδει τετραγώνῳ, ἔωντι δὲ αἱ 
πλευραὶ τῶν ὑπερβλημάτων τῷ ἴσῳ ἀλλάλαν ὑπερεχούσαι. 


The rectangle applied is παράβλημα. 


Squares. 


A square is τετράγωνον, a square on a straight line is a square 


— 


(erected) from it (ἀπό). The square on TH, τὸ ἀπὸ τᾶς VE τετράγωνον, 


THE TERMINOLOGY OF ARCHIMEDES. elxi 


is shortened into τὸ ἀπὸ τᾶς ΓΈ, or τὸ ἀπὸ TE simply. The square 
next in order to it (when there are a number of squares in a row) is 
τὸ Tap αὐτῷ τετράγωνον OF τὸ ἐχόμενον τετράγωνον. 

With reference to squares, a most important part is played by 
the word δύναμις πα the various parts of the verb δύναμαι. δύναμις 
expresses a square (literally a power); thus in Diophantus it is used 
throughout as the technical term for the square of the unknown 
quantity in an algebraical equation, i.e. for αὖ. In geometrical 
language it is the dative singular δυνάμει which is mostly used ; 
thus a straight line is said to be potentially equal, δυνάμει ἴσα, to a 
certain rectangle where the meaning is that the sgware on the straight 
line is equal to the rectangle ; similarly for the square.on BA is less 
than double the square on AK we have ἡ BA ἐλάσσων ἐστὶν ἢ διπλασίων 
δυνάμει τῆς AK. The verb δύνασθαι (with or without ἴσον) has the 
sense of being δυνάμει ἴσα, and, when δύνασθαι is used alone, it is 
followed by the accusative ; thus the square (on a straight line) is 
equal to the rectangle contained by... is (εὐθεῖα) ἴσον δύναται τῷ 
περιεχομένῳ ὑπό... ; let the square on the radius be equal to the 
rectangle BA, AZ, ἡ ἐκ τοῦ κέντρου δυνάσθω τὸ ὑπὸ τών BAZ, (the 
difference) by which the square on ZV is greater than the square on 
half the other diameter ᾧ μεῖζον δυνάται ἃ ZT τᾶς ἡμισείας τᾶς ἑτέρας 
διαμέτρου. 

A gnomon is γνώμων, and its breadth (πλάτος) is the breadth of 
each end; a gnomon of breadth equal to BI, γνώμων πλάτος ἔχων ἴσον 
τᾷ BI, (a gnomon) whose breadth is greater by one segment than the 
breadth of the gnomon last taken away οὗ πλάτος ἑνὶ τμάματι μεῖζον 


lal 4 “ Ν 3 a 3 / he 
τοῦ πλάτεος τοῦ TPO αὐτοῦ ἀφαιρουμένου γνωμονος. 


Polygons. 


A polygon is πολύγωνον, an equilateral polygon is ἰσόπλευρον, 
a polygon of an even number of sides or angles ἀρτιόπλευρον or 
ἀρτιόγωνον ; a polygon with all its sides equal except BA, AA, ἴσας 
ἔχον Tas πλευρὰς χωρὶς τῶν BAA; a polygon with its sides, excluding 
the base, equal and even in number τὰς πλευρὰς ἔχον χωρὶς τῆς βάσεως 
ἴσας καὶ ἀρτίους : an equilateral polygon the number of whose sides is 
measured by four πολύγωνον ἰσόπλευρον, οὗ ai πλευραὶ ὑπὸ τετράδος 
μετροῦνται, let the number of its sides be measured by four τὸ πλῆθος 
τῶν πλευρῶν μετρείσθω ὑπὸ τετράδος. A chiliagon χιλιάγωνον. 

The straight lines subtending two sides of the polygon (i.e. joining 
angles next but one to each other) αἱ ὑπὸ δύο πλευρὰς τοῦ πολυγώνου 


H. A. 1 


elxil INTRODUCTION. 


ὑποτείνουσαι, the straight line subtending one less than half the 
. ine. 4 ἃς ADs / ~ ε , 
number of the sides n ὑποτείνουσα τὰς μιᾷ ἐλάσσονας τῶν ἡμίσεων. 


Circles. 


A circle is κύκλος, the circle Ψ is 6 Ψ κύκλος or ὃ κύκλος ἐν ᾧ τὸ Ψ, 
let the given circle be that drawn below ἔστω ὃ δοθεὶς κύκλος ὃ 
ὑποκείμενος. 

The centre is κέντρον, the circumference περιφέρεια, the former 
word having doubtless been suggested by something stwck im and 
the latter by something, e.g. a cord stretched tight, carried round 
the centre as a fixed point and describing a circle with its other 
extremity. Accordingly περιφέρεια is used for a circular are as well 
as for the whole circumference ; thus the are BA is ἡ BA περιφέρεια, 
the (part of the) circumference of the circle cut off by the same 
(straight line) ἡ τοῦ κύκλου περιφέρεια ἡ ὑπὸ τῆς αὐτῆς ἀποτεμνομένη. 
Though the circumference of a circle is also sometimes called its 
perimeter (ἢ περίμετρος) in the treatises On the Sphere and Cylinder 
and on the Measwrement of a Circle, the word does not seem to have 
been used by Archimedes himself in this sense ; he speaks, however, 
in the Sand-reckoner of the perimeter of the earth (περίμετρος Tas yas). 

The radius is y ἐκ τοῦ κέντρου simply, and this expression 
without the article is used as a predicate as if it were one word ; 
thus the circle whose radius is ΘῈΣ is ὃ κύκλος ot ἐκ τοῦ κέντρου a 
OE; BE is a radius of the circle ἡ δὲ BE ἐκ τοῦ κέντρου ἐστὶ τοῦ κύκλου. 

A diameter is διάμετρος, the circle on AE as diameter ὃ περὶ 
διάμετρον τὴν AE κύκλος. 

For drawing a chord of a circle there is no special technical 
term, but we find such phrases as the following: ἐὰν eis τὸν κύκλον 
εὐθεῖα γραμμὴ ἐμπέσῃ if in a circle ὦ straight line be placed, and the 
chord is then the straight line so placed ἡ ἐμπεσοῦσα, or quite 
commonly 7 ἐν τῷ κύκλῳ (εὐθεῖα) simply. For the chord subtending 
one 656th part of the circumference of a circle we have the following 
interesting phrase, ἃ ὑποτείνουσα ἕν τμᾶμα διαιρεθείσας Tas τοῦ ABI 
κύκλου περιφερείας ἐς χνς΄. 

A segment of a circle is τμῆμα κύκλου ; sometimes, to distinguish 
it from a segment of a sphere, it is called a plane segment 
τμῆμα ἐπίπεδον. A semicircle is ἡμικύκλιον ; a segment less than a 
semicircle cut off by AB, τμῆμα ἔλασσον ἡμικυκλίου ὃ ἀποτέμνει 
ἡ AB. The segments on AE, EB (as bases) are ta ἐπὶ τῶν 
AE, EB τμήματα; but the semicircle on ZH as diameter is τὸ 


THE TERMINOLOGY OF ARCHIMEDES. elxili 


ἡμικύκλιον TO περὶ διάμετρον τὰν ZH or τὸ ἡμικύκλιον τὸ περὶ τὰν ΖΗ 
simply. The expression the angle of the semicircle, ἃ τοῦ ἡμικυκλίου 
(ywvia), is used of the (right) angle contained by the diameter and 
the arc (or tangent) at one extremity of it. 

A sector of a circle is τομεύς or, when it is necessary to 
distinguish it from what Archimedes calls a ‘solid sector,’ ἐπίπεδος 
τομεὺς κύκλου a plane sector of a circle. The sector including the 
right angle (at the centre) is ὃ τομεὺς ὃ τὰν ὀρθὰν γωνίαν περιέχων. 
Either of the radii bounding a sector is called a side of it, πλευρά ; 
each of the sectors (is) equal to the sector which has a side common 
(with it) ἕκαστος τῶν τομέων ἴσος τῷ κοινὰν ἔχοντι πλευρὰν τομεῖ ; a 
sector is sometimes regarded as described on one of the bounding 
radii as a side, thus similar sectors have been described on all (the 
straight lines) ἀναγεγραφάται ἀπὸ πασᾶν ὁμοίοι τομέες. 

Of polygons inscribed in or circumscribed about a circle ἐγγράφειν 
eis or ἐν and περιγράφειν περί (with acc.) are used; we also find 
περιγεγραμμένος used with the simple dative, thus τὸ περιγε- 
γραμμένον σχῆμα τῷ τομεῖ is the figure circumscribed to the sector. 
A polygon is said to be inscribed in a segment of w circle when 
the base of the segment is one side and the other sides subtend 
ares making up the circumference ; thus let a polygon be inscribed 
on AY in the segment ABI, ἐπὶ τῆς AT πολύγωνον ἐγγεγράφθω 
eis τὸ ABT τμῆμα. A regular polygon is said to be wnseribed in 
a sector when the two radii are two of the sides and the other sides 
are all equal to one another, and a similar polygon is said to be 
circumscribed about a sector when the equal sides are formed by the 
tangents to the arc which are respectively parallel to the equal 
sides of the inscribed polygon and the remaining two sides are the 
bounding radii produced to meet the adjacent tangents. Of a 
circle circwmscribed to a polygon περιλαμβάνειν is also used; thus 
πολύγωνον κύκλος περιγεγραμμένος περιλαμβανέτω περὶ TO αὐτὸ κέντρον 
γινόμενος, as we might say let ὦ circumscribed circle be drawn with 
the same centre going round the polygon. Similarly the circle ABTA 
containing the polygon ὃ ABTA κύκλος ἔχων τὸ πολύγωνον. 

When a polygon is inscribed in a circle, the segments left over 
between the sides of the polygon and the subtended arcs are 
περιλειπόμενα τμήματα; when a polygon is circumscribed to the 
circle, the spaces between the two are variously called τὰ περι- 
λειπόμενα τῆς περιγραφῆς τμήματα, τὰ περιλειπόμενα σχήματα, τὰ 
περιλείμματα Or τὰ ἀπολείμματα. 


clxiv INTRODUCTION. 


Spheres, etc. 


In connexion with a sphere (σφαῖρα) a number of terms are 
used on the analogy of the older and similar terms connected with 
the circle. Thus the centre is κέντρον, the radius ἡ ἐκ τοῦ κέντρου, 
the diameter y διάμετρος. Two segments, τμήματα σφαίρας or 
τμήματα σφαιρικά, are formed when a sphere is cut by a plane; 
a hemisphere is ἡμισφαίριον ; the segment of the sphere at 1, τὸ κατὰ τὸ 
Τ' τμῆμα τῆς σφαίρας ; the segment on the side of ABT, τὸ ἀπὸ ABT 
τμῆμα; the segment including the circumference BAA, τὸ κατὰ τὴν BAA 
περιφέρειαν τμῆμα. The curved surface of a sphere or segment 
is ἐπιφάνεια ; thus of spherical segments bounded by equal surfaces the 
hemisphere is greatest is τῶν τῇ ion ἐπιφανείᾳ περιεχομένων σφαιρικῶν 
τμημάτων μεῖζόν ἐστι τὸ ἡμισφαίριον. The terms base (βάσις), vertex 
(κορυφή) and height (ὕψος) are also used with reference to a segment 
of a sphere. 

Another term borrowed from the geometry of the circle is the 
word sector (τομεύς) qualified with the adjective στερεός (solid). 
A solid sector (τομεὺς στερεός) is defined by Archimedes as the 
figure bounded by a cone which has its vertex at the centre of 
a sphere and the part of the surface of the sphere within the cone. 
The segment of the sphere included wn the sector is τὸ τμῆμα τῆς 
σφαίρας τὸ ἐν τῷ τομεῖ OF TO κατὰ TOV τομέα. 

A great circle of a sphere is ὃ μέγιστος κύκλος τῶν ἐν τῇ σφαίρᾳ 
and often ὃ μέγιστος κύκλος alone. 

Let a sphere be cut by a plane not through the centre τετμήσθω 
σφαῖρα μὴ διὰ τοῦ κέντρου ἐπιπέδῳ ; a sphere cut by a plane through 
the centre in the circle EZH®, σφαῖρα ἐπιπέδῳ τετμημένη διὰ τοῦ 


κέντρου κατὰ τὸν ΕΖΗΘ κύκλον. 


Prisms and pyramids, 

A prism is πρῖσμα, a pyramid πυραμίς. As usual, ἀναγράφειν ἀπό 
is used of describing a prism or pyramid on a rectilineal figure 
as base; thus let a prism be described on the rectilineal figure 
(as base) ἀναγεγράφθω ἀπὸ τοῦ εὐθυγράμμου πρῖσμα, on the polygon 
circumscribed about the circle A let a pyramid be set wp ἀπὸ τοῦ περὶ 
τὸν A κύκλον περιγεγραμμένου πολυγώνου πυραμὶς ἀνεστάτω ἀναγεγραμ- 
μένη. A pyramid with an equilateral base ABT is πυραμὶς ἰσόπλευρον 
ἔχουσα βάσιν τὸ ABY. 

The surface is, as usual, ἐπιφάνεια and, when any particular face 
or a base is excluded, some qualifying phrase has to be used. 


THE TERMINOLOGY OF ARCHIMEDES. elxv 


Thus the surface of the prism consisting of the parallelograms 
(ie. excluding the bases) ἡ ἐπιφάνεια τοῦ πρίσματος ἡ ἐκ τῶν 
παραλληλογράμμων συγκειμένη ; the sunface (of a pyramid) excluding 
the base or the triangle AET, ἡ ἐπιφάνεια χωρὶς τῆς βάσεως or τοῦ 
AED τριγώνου. 

The triangles bounding the pyramid τὰ περιέχοντα τρίγωνα τὴν 
πυραμίδα (as distinct from the base, which may be polygonal). 


Cones and solid rhombi. 


The Elements of Euclid only introduce right cones, which are 
simply called cones without the qualifying adjective. A cone is 
there defined as the surface described by the revolution of a right- 
angled triangle about one of the sides containing the right angle. 
Archimedes does not define a cone, but generally describes a right 
cone as an isosceles cone (κῶνος ἰσοσκελής), though once he calls it 
right (ὀρθός). J. H. T. Miiller rightly observes that the term 
isosceles applied to a cone was suggested by the analogy of the 
isosceles triangle, but I doubt whether such a cone was thought of 
(as he supposes) as one which could be described by making an 
isosceles triangle revolve about the perpendicular from the vertex 
on the base; it seems more natural to connect it with the use of 
the word side (πλευρά) by which Archimedes designates a generator 
of the cone, a right cone being thus directly regarded as a cone having 
all its legs equal. The latter supposition would also accord better 
with the term scalene cone (κῶνος σκαληνός) by which Apollonius 
denotes an oblique circular cone; such a cone could not of course 
be described by the revolution of a scalene triangle. An oblique 
circular cone is simply a cone for Archimedes, and he does not 
define it; but, while he speaks of finding a cone with a given 
vertex and passing through every point on a given ‘section of an 
acute-angled cone’ [ellipse], he regards the finding of the cone as 
being equivalent to finding the circular sections, and we may 
therefore conclude that he would have defined the cone in 
practically the same way as Apollonius does, namely as the surface 
described by a straight line always passing through a fixed point 
and moving round the circumference of any circle not in the same 
plane with the point. 

The vertex of a cone is, as usual, κορυφή, the base βάσις, the axis 
ἄξων and the height ὕψος ; the cones are of the same height εἰσὶν οἱ 
κῶνοι ὑπὸ TO αὐτὸ ὕψος. A generator is called a side (πλευρά); of α 


clxvi INTRODUCTION. 


cone be cut by a plane meeting all the generators of the cone εἴ κα 
κῶνος ἐπιπέδῳ τμαθῇ συμπίπτοντι πάσαις ταῖς TOD κώνου πλευραῖς. 

The surface of the cone excluding the base ἡ ἐπιφάνεια τοῦ κώνου 
χωρὶς τῆς βάσεως ; the conical surface between (two generators) AA, AB, 
κωνικὴ ἐπιφάνεια ἡ μεταξὺ τῶν ΑΔΒ. 

There is no special name for what we call ἃ frustwm of a cone 
or the portion intercepted between two planes parallel to the base ; 
the surface of such a frustum is simply the surface of the cone 
between the parallel planes ἢ ἐπιφάνεια τοῦ κώνου μεταξὺ τῶν 
παραλλήλων ἐπιπέδων. 

A curious term is segment of a cone (ἀπότμαμα κώνου), which is 
used of the portion of any circular cone, right or oblique, cut off 
towards the vertex by any plane which makes an elliptic and not a 
circular section. With reference to a segment of a cone the axis 
(ἄξων) is defined as the straight line drawn from the vertex of the 
cone to the centre of the elliptic base. 

As usual, ἀναγράφειν ἀπὸ is used of describing a cone on a circle 
as base. Similarly, a very common phrase is ἀπὸ τοῦ κύκλου κῶνος 
ἔστω let there be a cone on the circle (as base). 

A solid rhombus (ῥόμβος στερεός) is the figure made up of two 
cones having their base common, their vertices on opposite sides of 
it, and their axes in one straight line. A rhombus made up of 
isosceles cones ῥόμβος ἐξ ἰσοσκελῶν κώνων συγκείμενος, and the two 
cones are spoken of as the cones bounding the rhombus ot κῶνοι ot 
περιέχοντες TOV ῥόμβον. 


Cylinders. 


A right cylinder is κύλινδρος ὀρθός, and the following terms 
apply to the cylinder as to the cone: base βάσις, one base or the 
other ἡ ἑτέρα βάσις, of which the circle AB is a base and TA opposite 
to it οὗ βάσις μὲν ὃ AB κύκλος, ἀπεναντίον δὲ ὁ TA; axis ἄξων, height 
ὕψος, generator πλευρά. The cylindrical surface cut off by (two 
generators) AT, BA, ἡ ἀποτεμνομένη κυλινδρικὴ ἐπιφάνεια ὑπὸ τῶν AT, 
ΒΔ; the surface of the cylinder adjacent to the circumference ABY, ἡ 
ἐπιφάνεια τοῦ κυλίνδρου ἡ κατὰ τὴν ABI περιφέρειαν denotes the 
surface of the cylinder between the two generators drawn through 
the extremities of the arc. 

A frustum of a cylinder topos κυλίνδρου is a portion of a 
cylinder intercepted between two parallel sections which are elliptic 
and not circular, and the axis (ἄξων) of it is the straight line 


THE TERMINOLOGY OF ARCHIMEDES. elxvil 


joining the centres of the two sections, which is in the same straight 
line with the axis of the cylinder. 


Conic Sections, 


General terms are κωνικὰ στοιχεῖα, elements of conics, τὰ κωνικά 
(the theory of) conics. Any conic section κώνου τομὴ ὅποιαοῦν. 
Chords are simply εὐθείαι ἐν τᾷ τοῦ κώνου τομᾷ ἀγμέναι. Archimedes 
never uses the word awis (aéwv) with reference to a conic ; the axes 
are with him diameters (διάμετροι), and διάμετρος, when it has 
reference to a complete conic, is used in this sense exclusively. A 
tangent is ἐπιψαύουσα or ἐφαπτομένη (with gen.). 

The separate conic sections are still denoted by the old names ; 
a parabola is a section of a right-angled cone ὀρθογωνίου κώνου τομή, 
a hyperbola a section of an obtuse-angled cone ἀμβλυγωνίου κώνου 
τομή, and an ellipse a section of an acute-angled cone ὀξυγωνίου κώνου 


τομή. 


The parabola. 


Only the axis of a complete parabola is called a diameter, and 
the other diameters are simply lines parallel to the diameter. Thus 
parallel to the diameter or itself the diameter is παρὰ τὰν διάμετρον ἢ 
αὐτὰ διάμετρος; AZ is parallel to the diameter ἃ AZ παρὰ τὰν 
διάμετρόν ἐστι. Once the term principal or original (diameter) is 
used, ἀρχικά (sc. διάμετρος). 

A. segment of a parabola is τμῆμα, which is more fully described 
as the segment bounded by a straight line and a section of a right- 
angled cone τμᾶμα τὸ περιεχόμενον ὑπό τε εὐθείας Kal ὀρθογωνίου κώνου 
τομᾶς. The word διάμετρος is again used with reference to ἃ 
segment of a parabola in the sense of our word awis; Archimedes 
defines the diameter of any segment as the line bisecting all the 
straight lines (chords) drawn parallel to its base τὰν δίχα τέμνουσαν 
τὰς εὐθείας πάσας Tas παρὰ τὰν βάσιν αὐτοῦ ἀγομένας. 

The part of a parabola included between two parallel chords is 
called a frustwm τόμος (ἀπὸ ὀρθογωνίου κώνου Tomas ἀφαιρούμενος), 
the two chords are its lesser and greater base (ἐλάσσων and μείζων 
βάσις) respectively, and the line joining the middle points of the 
two chords is the diameter (διάμετρος) of the frustum. 

What we call the Jatus rectum of a parabola is in Archimedes 
the line which is double of the line drawn as far as the axis ἃ διπλασία 
τᾶς μέχρι τοῦ ἄξονος. In this expression the awis (aéwv) is the axis 


elxvill INTRODUCTION. 


of the right-angled cone from which the curve was originally derived 
by means of a section perpendicular to a generator*. Or, again, the 
equivalent of our word parameter (παρ᾽ ἃν δυνάνται at ἀπὸ τᾶς Topas) 
is used by Archimedes as by Apollonius, meaning the straight line 
to which the rectangle which has its breadth equal to the abscissa 
of a point and is equal to the square of the ordinate must be 
applied as base. The full phrase states that the ordinates have 
their squares equal to the rectangles applied to the line equal to N (or 
the parameter) which have as their breadth the lines which they (the 
ordinates) cut off from AZ (the diameter) towards the extremity A, 
δυνάνται τὰ παρὰ τὰν ἴσαν τᾷ N παραπίπτοντα πλάτος ἔχοντα, as αὐταὶ 
ἀπολαμβάνοντι ἀπὸ τᾶς ΔΖ ποτὶ τὸ Δ πέρας. 

Ordinates are the lines drawn from the section to the diameter 
(of the segment) parallel to the base (of the segment) ai ἀπὸ τᾶς Topas 
ἐπὶ τὰν AZ ἀγομέναι παρὰ τὰν AE, or simply ai ἀπὸ τᾶς touds. Once 
also the regular phrase drawn ordinate-wise τεταγμένως κατηγμένη 15 
used to describe an ordinate, as in Apollonius. 


The hyperbola. 


What we call the asymptotes (ai ἀσύμπτωτοι in Apollonius) are 
in Archimedes the lines (approaching) nearest to the section of the 
obtuse-angled cone ai ἔγγιστα τᾶς τοῦ ἀμβλυγωνίου κώνου τομάς. 

The centre is not described as such, but it is the point at which 
the lines nearest (to the curve) meet τὸ σαμεῖον, καθ᾽ ὃ ai ἔγγιστα 
συμπιίιπτοντι. 

This is ὦ property of the sections of obtuse-angled cones τοῦτο γάρ 


“ na ’ a , 
ἐστιν ἐν ταῖς Tov ἀμβλυγωνίου κώνου τομαῖς σύμπτωμα. 


The ellipse. 


The major and minor axes are the greater and lesser diameters 
μείζων and ἐλάσσων διάμετρος. Let the greater diameter be AT, 
διάμετρος δὲ (αὐτᾶς) ἃ μὲν μείζων ἔστω ἐφ᾽ ἃς τὰ A,T. The rectangle 
contained by the diameters (axes) τὸ περιεχόμενον ὑπὸ τᾶν διαμέτρων. 
One axis is called conjugate (συζυγής) to the other: thus let the 
straight line N be equal to half of the other diameter which is 
conjugate to AB, ἁ δὲ Ν εὐθεῖα ἴσα ἔστω τᾷ ἡμισείᾳ τᾶς ἑτέρας διαμέτρου, 
ἅ ἐστι συζυγὴς τᾷ ΑΒ. 

The centre is here κέντρον. 


* Cf. Apollonius of Perga, pp. xxiv, xxv. 


THE TERMINOLOGY OF ARCHIMEDES. clxix 


Conoids and Spheroids., 


There is a remarkable similarity between the language in which 
Archimedes describes the genesis of his solids of revolution and that 
used by Euclid in defining the sphere. Thus Euclid says: when, the 
diameter of a semicircle remaining fixed, the semicircle revolves and 
returns to the same position from which it began to move, the included 
Jigure is a sphere σφαῖρά ἐστιν, ὅταν ἡμικυκλίου μενούσης τῆς διαμέτρου 
περιενεχθὲν τὸ ἡμικύκλιον εἰς τὸ αὐτὸ πάλιν ἀποκατασταθῇ, ὅθεν ἤρξατο 
φέρεσθαι, τὸ περιληφθὲν σχῆμα; and he proceeds to state that the 
axis of the sphere is the fixed straight line about which the semicircle 
turns ἄξων δὲ τῆς σφαίρας ἐστὶν ἡ μένουσα εὐθεῖα, περὶ ἣν τὸ ἡμικύκλιον 
στρέφεται. Compare with this e.g. Archimedes’ definition of the 
right-angled conoid (paraboloid of revolution): if ὦ section of a 
right-angled cone, with its diameter (axis) remaining fixed, revolves 
and returns to the position from which it started, the figure included 
by the section of the right-angled cone is called a right-angled conoid, 
and its axis is defined as the diameter which has remained fixed, 
εἴ Ka ὀρθογωνίου κώνου τομὰ μενούσας τᾶς διαμέτρου περιενεχθεῖσα 
ἀποκατασταθῇ πάλιν, ὅθεν ὥρμασεν, τὸ περιλαφθὲν σχῆμα ὑπὸ τᾶς τοῦ 
ὀρθογωνίου κώνου τομᾶς ὀρθογώνιον κωνοειδὲς καλείσθαι, καὶ ἄξονα 
μὲν αὐτοῦ τὰν μεμενακοῦσαν διάμετρον καλείσθαι, and it will be seen 
that the several phrases used are practically identical with those of 
Euclid, except that ὥρμασεν takes the place of ἤρξατο φέρεσθαι; and 
even the latter phrase occurs in Archimedes’ description of the 
genesis of the spiral later on. 

The words conoid κωνοειδὲς (σχῆμα) and spheroid σφαιροειδὲς 
(σχῆμα) are simply adapted from κῶνος and σφαῖρα, meaning that 
the respective figures have the appearance (εἶδος) of, or resemble, 
cones and spheres; and in this respect the names are perhaps more 
satisfactory than paraboloid, hyperboloid and ellipsoid, which can 
only be said to resemble the respective conics in a different sense. 
But when xwvoedés is qualified by the adjective right-angled 
ὀρθογώνιον to denote the paraboloid of revolution, and by ἀμβλυ- 
γώνιον obtuse-angled to denote the hyperboloid of revolution, the 
expressions are less logical, as the solids do not resemble right- 
angled and obtuse-angled cones respectively; in fact, since the 
angle between the asymptotes of the generating hyperbola may be 
acute, a hyperboloid of revolution would in that case more resemble 
an acute-angled cone. The terms right-angled and obtuse-angled 


elxx INTRODUCTION. 


were merely transferred to the conoids from the names for the 
respective conics without any more thought of their meaning. 

It is unnecessary to give separately the definition of each 
conoid and spheroid; the phraseology is in all cases the same 
as that given above for the paraboloid. But it may be remarked 
that Archimedes does not mention the conjugate axis of a hyperbola 
or the figure obtained by causing a hyperbola to revolve about that 
axis ; the conjugate axis of a hyperbola first appears in Apollonius, 
who was apparently the first to conceive of the two branches of a 
hyperbola as one curve. Thus there is only one obtuse-angled 
conoid in Archimedes, whereas there are two kinds of spheroids 
according as the revolution takes place about the greater diameter 
(axis) or lesser diameter of the generating section of an acute- 
angled cone (ellipse); the spheroid is in the former case oblong 
(παραμᾶκες σφαιροειδές) and in the latter case flat (ἐπιπλατὺ 
σφαιροειδές). 

A special feature is, however, to be observed in the description 
of the obtuse-angled conoid (hyperboloid of revolution), namely that 
the asymptotes of the hyperbola are supposed to revolve about the 
axis at the same time as the curve, and Archimedes explains that 
they will include an isosceles cone (κῶνον ἰσοσκελέα περιλαψούνται), 
which he thereupon defines as the cone enveloping the conoid 
(περιέχων τὸ κωνοειδές). Also in a spheroid the term diameter 
(διάμετρος) is appropriated to the straight line drawn through 
the centre at right angles to the awis (ἃ διὰ τοῦ κέντρου ποτ᾽ ὀρθὰς 
ἀγομένα τῷ agovr). The centre of a spheroid is the middle point of 
the axis τὸ μέσον τοῦ a€ovos. 

The following terms are used of all the conoids and spheroids. 
The vertex (κορυφή) is the point at which the axis meets the surface τὸ 
σαμεῖον, καθ᾽ ὃ ἁπτέται 6 ἀξων τᾶς ἐπιφανείας, the spheroid having of 
course two vertices. A segment (τμᾶμαλ) is a part cut off by a plane, 
and the base (βάσις) of the segment is defined as the plane ( figure) 
included by the section of the conoid (or spheroid) in the cutting 
plane τὸ ἐπίπεδον τὸ περιλαφθὲν ὑπὸ τᾶς τοῦ κωνοειδέος (Or σφαιροειδέος) 
τομάς ἐν τῷ ἀποτέμνοντι ἐπιπέδῳ. The vertex of a segment is the point 
at which the tangent plane parallel to the base of the segment meets 
the surface, τὸ capetov, καθ᾽ ὃ ἁπτέται τὸ ἐπίπεδον τὸ ἐπιψαῦον (τοῦ 
κωνοειδέος). The axis (agwv) of a segment is differently defined for 
the three surfaces ; (a) in the paraboloid it is the straight line cut off 
within the segment from the line drawn through the vertex of the 


THE TERMINOLOGY OF ARCHIMEDES. elxxi 


segment parallel to the axis of the conoid ἃ ἐναπολαφθεῖσα εὐθεῖα ἐν τῷ 
τμάματι ἀπὸ τᾶς ἀχθείσας διὰ τᾶς κορυφᾶς τοῦ τμάματος παρὰ TOV 
ἄξονα τοῦ κωνοειδέος, (6) in the hyperboloid it is the straight line cut 
off within the segment from the line drawn through the vertex of the 
segment and the vertex of the cone enveloping the conoid ἀπὸ tas 
ἀχθείσας διὰ Tas κορυφᾶς τοῦ τμάματος καὶ τᾶς κορυφᾶς τοῦ κώνου TOD 
περιέχοντος τὸ κωνοειδές, (6) in the spheroid it is the part similarly 
cut off from the straight line joining the vertices of the two segments 
into which the base divides the spheroid, ἀπὸ τᾶς εὐθείας τᾶς τὰς 
κορυφὰς αὐτῶν (τῶν τμαμάτων) ἐπιζευγνυούσας. 

Archimedes does not use the word centre with respect to the 
hyperboloid of revolution, but calls the centre the vertex of the 
enveloping cone. Also the axis of a hyperboloid or a segment is 
only that part of it which is within the surface. The distance 
between the vertex of the hyperboloid or segment and the vertex 
of the enveloping cone is the line adjacent to the axis ἃ ποτεοῦσα 
τῷ ἀξονι. ᾿ 

The following are miscellaneous expressions. Zhe part inter- 
cepted within the conoid of the intersection of the planes ἃ ἐναπο- 
λαφθεῖσα ἐν τῷ κωνοειδεῖ Tas γενομένας τομᾶς τῶν ἐπιπέδων, (the plane) 
will have cut the spheroid through its axis τετμακὸς ἐσσείται τὸ 
σφαιροειδὲς διὰ τοῦ ἀξονος, so that the section it makes will be a 
conic section ὥστε τὰν τομὰν ποιήσει κώνου Tomar, let two segments be 
cut off in any manner ἀποτετμάσθω δύο τμάματα ws ἔτυχεν or by 
planes drawn in any manner ἐπιπέδοις ὁπωσοῦν aypévoss. 

Half the spheroid τὸ ἁμίσεον tod σφαιροειδέος, half the line 
joining the vertices of the segments (of a spheroid), i.e. what we should 
call a semi-diameter, a ἡμισέα αὐτᾶς tas ἐπιζευγνυούσας τὰς κορυφὰς 


“ , 
τῶν τμαμαάτων. 


The spiral. 


We have already had, in the conoids and spheroids, instances of 
the evolution of figures by the motion of curves about an axis. The 
same sort of motion is used for the construction of solid figures 
inscribed in and circumscribed about a sphere, a circle and an 
inscribed or circumscribed polygon being made to revolve about 
a diameter passing through an angular point of the polygon and 
dividing it and the circle symmetrically, In this case, in Archimedes’ 
phrase, the angular points of the polygon will move along the circwm- 
Jerences of circles, ai γωνίαι κατὰ κύκλων περιφερειῶν ἐνεχθήσονται (or 


elxxl INTRODUCTION. 


οἰσθήσονται) and the sides will move on certain cones, or on the surface 
of a cone κατά τινων κώνων ἐνεχθήσονται or κατ᾽ ἐπιφανείας κώνου ; and 
sometimes the angular points or the points of contact of the sides of 
a circumscribed polygon are said to describe circles γράφουσι κύκλους. 
The solid figure so formed is τὸ γενηθὲν στερεὸν σχῆμα, and let the 
sphere by its revolution make a figure περιενεχθεῖσα ἡ σφαῖρα ποιείτω 
σχῆμά τι. 

For the construction of the spiral, however, we have a new 
element introduced, that of time, and we have two different uniform 
motions combined ; if a straight line in a plane turn uniformly 
about one extremity which remains fixed, and return to the position 
Jrom which ut started and if, at the same time as the line is revolving, 
a point move at a uniform rate along the line starting from the fixed 
extremity, the point will desoribe a spiral in the plane, εἴ κα εὐθεῖα... ἐν 
ἐπιπέδῳ... μένοντος τοῦ ἑτέρου πέρατος αὐτᾶς ἰσοταχέως περιενεχθεῖσα 
ἀποκατασταθῇ πάλιν, ὅθεν ὥρμασεν, ἅμα δὲ τᾷ γραμμᾷ περιαγομένᾳ 
φερήται τι σαμεῖον ἰσοταχέως αὐτὸ ἑαυτῷ κατὰ τᾶς εὐθείας ἀρξάμενον ἀπὸ 
τοῦ μένοντος πέρατος, τὸ σαμεῖον ἕλικα γράψει ἐν τῷ ἐπιπέδῳ. 

The spiral (described) in the first, second, or any turn is & ἕλιξ ἃ ἐν 
τᾷ πρώτᾳ, δευτέρᾳ, OY ὁποιᾳοῦν περιφορᾷ γεγραμμένα, and the turns 
other than any particular ones are the other spirals at ἀλλαι ἕλικες. 

The distance traversed by the point along the line in any time is 
a εὐθεῖα a διανυσθεῖσα, and the times in which the point moved over the 
distances οἱ χρόνοι, ἐν οἷς τὸ σαμεῖον Tas γραμμὰς ἐπορεύθη ; in the tume 
in which the revolving line reaches AT from AB, ἐν ᾧ χρόνῳ ἃ περιαγομένα 
γραμμὰ ἀπὸ τᾶς AB ἐπὶ τὰν AT ἀφικνείται. 

The origin of the spiral is ἀρχὰ τᾶς ἕλικος, the intial line ἀρχὰ τᾶς 
περιφορᾶς. The distance described by the point along the line in 
the first complete revolution is εὐθεῖα πρώτα (first distance), that 
described during the second revolution the second distance εὐθεῖα 
δευτέρα, and so on, the distances being called by the number of the 
revolutions ὁμωνύμως ταῖς περιφοραῖς. The first area, χωρίον πρῶτον, 
is the area bounded by the spiral described in the first revolution and 
by the ‘first distance’ τὸ χωρίον τὸ περιλαφθὲν ὑπό τε Tas ἕλικος τᾶς ἐν 
τᾷ πρώτᾳ περιφορᾷ γραφείσας καὶ τᾶς εὐθείας, a ἐστιν πρώτα : the second 
area is that bounded by the spiral in the second turn and the ‘second 
distance,’ and so on. The area added by the spiral in any turn is τὸ 
χωρίον τὸ ποτιλαφθὲν ὑπὸ τᾶς ἕλικος ἔν τινι περιφορᾷ. 

The first circle, κύκλος πρῶτος, is the circle described with the 
‘first distance’ as radius and the origin as centre, the second circle 


THE TERMINOLOGY OF ARCHIMEDES. elxxili 


that with the origin as centre and twice the ‘first distance’ as 
radius, and so on. 

Together with as many times the whole of the circumference of the 
circle as (is represented by) the number less by one than (that of) 
the revolutions μεθ᾽ ὅλας τᾶς τοῦ κύκλου περιφερείας τοσαυτάκις Aap- 
βανομένας, ὅσος ἐστὶν ὃ ἑνὶ ἐλάσσων ἀριθμὸς τᾶν περιφορᾶν, the cirele 
called by the number corresponding to that of the revolutions ὃ κύκλος 
6 κατὰ τὸν αὐτὸν ἀριθμὸν λεγόμενος ταῖς περιφοραῖς. 

With reference to any radius vector, the side which is in the 
direction of the revolution is forward τὰ προαγούμενα, the other 
backward τὰ ἑπόμενα. 


Tangents, etc. 


Though the word ἅπτομαι is sometimes used in Archimedes of a 
line touching a curve, its general meaning is not to touch but simply 
to meet; e.g. the axis of a conoid or spheroid meets (ἅπτεται) the 
surface in the vertex. (The word is also often used elsewhere than 
in Archimedes of points /ying on a locus ; e.g. in Pappus, p. 664, the 
point will lie on a straight line given in position ἅψεται τὸ σημεῖον 
θέσει δεδομένης εὐθείας.) 

To touch a curve or surface is generally ἐφάπτεσθαι or ἐπιψαύειν 
(with gen.). A tangent is ἐφαπτομένη or ἐπιψαύουσα (sc. εὐθεῖα) and 
a tangent plane érupatov ἐπίπεδον. Let tangents be drawn to the circle 
ABD, τοῦ ΑΒΤ' κύκλου ἐφαπτόμεναι ἤχθωσαν; if straight lines be drawn 
touching the circles ἐὰν. ἀχθῶσίν τινες ἐπιψαύουσαι τῶν κύκλων. The 
full phrase of touching without cutting is sometimes found in 
Archimedes; if a plane touch (any of) the conoidal figures 
without cutting the conoid εἴ κα τῶν κωνοειδέων σχημάτων ἐπίπεδον 
ἐφαπτήται μὴ τέμνον τὸ κωνοειδές. The simple word ψαύειν is 
occasionally used (participially), the tangent planes τὰ ἐπίπεδα τὰ 
ψαύοντα. . 

To touch αὐ a point is expressed by κατά (with acc.) ; the points 
at which the sides...touch (or meet) the circle σημεῖα, καθ᾽ ἃ ἅπτονται 
τοῦ κύκλου ai πλευραί.... Let them touch the circle at the middle 
points of the circumferences cut off by the sides of the inscribed 
polygon ἐπιψαυέτωσαν τοῦ κύκλου κατὰ μέσα τῶν περιφερειῶν τῶν 
ἀποτεμνομένων ὑπὸ τοῦ ἐγγεγραμμένου πολυγώνου πλευρών. 

The distinction between ἐπιψαύειν and ἅπτομαι is well brought 
out in the following sentence; but that the planes touching the 
spheroid meet its surface at one point only we shall prove ὅτι δὲ 


clxxiv INTRODUCTION. 


τὰ ἐπιψαύοντα ἐπίπεδα τοῦ σφαιροειδέος καθ᾽ ἕν μόνον ἁπτόνται σαμεῖον 
τᾶς ἐπιφανείας αὐτοῦ δειξοῦμες. 

The point of contact ἡ ἁφή. 

Tangents drawn from (a point) ἀγμέναι ἀπό; we find also the 
elliptical expression ἀπὸ τοῦ Ξὶ ἐφαπτέσθω ἡ OT, let ΟΞΠ be the 
tangent from &, where, in the particular case, = is on the circle. 


Constructions. 


The richness of the Greek language in expressions for con-- 
structions is forcibly illustrated by the variety of words which 
may be used (with different shades of meaning) for drawing a 
line. Thus we have in the first place ayw and the compounds 
διάγω (of drawing a line through a figure, with εἰς or ἐν following, 
of producing a plane beyond a figure, or of drawing a line in ἃ 
plane), κατάγω (used of drawing an ordinate down from a point on 
a conic), προσάγω (of drawing a line ¢o meet another). As an 
alternative to προσάγω, προσβάλλω is also used; and προσπίπτω 
may take the place of the passive of either verb. To produce is 
ἐκβάλλω, and the same word is also used of a plane drawn through a 
point or through a straight line; an alternative for the passive 1s 
supplied by ἐκπίπτω. Moreover πρόσκειμαι is an alternative word 
for being produced (literally being added). 

In the vast majority of cases constructions are expressed by the 
elegant use of the perfect imperative passive (with which may be 
classed such forms as γεγονέτω from γίγνομαι, ἔστω from εἰμί, and 
κείσθω from κεῖμαι), or occasionally the aorist imperative passive. 
The great variety of the forms used will be understood from the 
following specimens. Let BI be made (or supposed) equal to A, 
κείσθω τῷ Δ ἴσον τὸ BY; let it be drawn ἤχθω, let ὦ straight line be 
drawn in it (a chord of a circle) διήχθω τις εἰς αὐτὸν εὐθεῖα, let KM be 
drawn equal to... ton κατήχθω ἡ KM, let it be joined ἐπεζεύχθω, let 
KA be drawn to meet προσβεβλήσθω ἡ KA, let them be produced 
ἐκβεβλήσθωσαν, suppose them found εὑρήσθωσαν, let a circle be set out 
ἐκκείσθω κύκλος, let it be taken εἰλήφθω, let K, H be taken ἔστωσαν 
εἰλημμέναι at K, H, let a circle Ψ be taken χελάφθω κύκλος ἐν ᾧ τὸ Ψ, let 
it be cut τετμήσθω, let it be divided διαιρήσθω (διῃρήσθω) ; let one cone be 
cut by a plane parallel to the base and produce the section EZ, τμηθήτω 6 
ἕτερος κῶνος ἐπιπέδῳ παραλλήλῳ τῇ βάσει καὶ ποιείτω τομὴν τὴν BZ, let 
TZ be cut off ἀπολελάφθω ἁ TZ; let (such an angle) be left and let rt 
be NHI, λελείφθω καὶ ἔστω ἡ ὑπὸ NHI, let a figure be made γεγενήσθω 


THE TERMINOLOGY OF ARCHIMEDES. elxxv 


σχῆμα, let the sector be made ἔστω γεγενημένος ὁ τομεύς, let cones be 
described on the circles (as bases) ἀναγεγράφθωσαν ἀπὸ τῶν κύκλων 
κῶνοι, ἀπὸ TOD κύκλου κῶνος ἔστω, let it be inscribed or circumscribed 
ἐγγεγράφθω (or ἐγγεγραμμένον ἔστω), περιγεγράφθω ; let an area (equal 
to that) of AB be applied to AH, παραβεβλήσθω rapa τὰν AH τὸ χωρίον 
τοῦ AB; let a segment of a circle be described on OK, ἐπὶ τῆς OK 
κύκλου τμῆμα ἐφεστάσθω, let the circle be completed ἀναπεπληρώσθω ὃ 
κύκλος, let NE (a parallelogram) be completed συμπεπληρώσθω τὸ NE, 
let it be made πεποιήσθω, let the rest of the construction be the same as 
before τὰ ἄλλα κατεσκευάσθω τὸν αὐτὸν τρόπον τοῖς πρότερον. Swppose 
it done γεγονέτω. 

Another method is to use the passive imperative of νοέω (/et it be 
conceived). Let straight lines be conceived to be drawn νοείσθωσαν 
εὐθεῖαι nypevat, let the sphere be conceived to be cut νοείσθω ἡ σφαῖρα 
τετμημένη, let a figure (generated) from the inscribed polygon be 
conceived as inscribed in the sphere ἀπὸ τοῦ πολυγώνου τοῦ ἐγγραφο- 
μένου νοείσθω τι εἰς τὴν σφαῖραν ἐγγραφὲν σχῆμα. Sometimes the 
participle for drawn is left out; thus ἀπ᾽ αὐτοῦ νοείσθω ἐπιφάνεια Let 
a surface be concewved (generated) from tt. 

The active is much more rarely used; but we find (1) ἐὰν with 
subjunctive, 1f we cut ἐὰν τέμωμεν, if we draw ἐὰν ἀγάγωμεν, if you 
produce ἐὰν ἐκβαλῇς ; (2) the participle, ἐέ is possible to inscribe...and 
(ultimately) to leave δυνατόν ἐστιν ἐγγράφοντα...λείπειν, if we con- 
tinually cirewmscribe polygons, bisecting the remaining circumferences 
and drawing tangents, we shall (ultimately) leave ἀεὶ δὴ περιγράφοντες 
πολύγωνα δίχα τεμνομένων TOV περιλειπομένων περιφερειῶν καὶ ἀγομένων 
ἐφαπτομένων λείψομεν, it is possible, if we take the area..., to inscribe 
λαβόντα (or λαμβάνοντα) τὸ Xwpiov...dvvarov ἐστιν... ἐγγράψαι ; (3) the 
first person singular, J take two straight lines λαμβάνω δύο εὐθείας, 
1] took a straight line ἔλαβόν twa εὐθεῖαν ; 7 draw ΘΜ from ® parallel 
to AZ, ἄγω ἀπὸ τοῦ Θ τὰν ΘΜ παράλληλον τᾷ AZ, having drawn TK 
perpendicular, I cut off AK equal to TK ἀγαγὼν κάθετον τὰν TK τᾷ 
TK ἴσαν ἀπέλαβον τὰν AK, 7 inscribed a solid figure...and cirewm- 
scribed another ἐνέγραψα σχῆμα oTepeov...kat ἄλλο περιέγραψα. 

The genitive of the passive participle is used absolutely, 
εὑρεθέντος δή it being supposed found, ἐγγραφέντος δή (the figure) 
being inscribed. 

To make a figure similar to one (and equal to another) δμοιώσαι, 


to find experimentally ὀργανικῶς λαβεῖν, to cut into unequal parts εἰς 
ἄνισα τέμνειν. 


clxxvi INTRODUCTION. 


Operations (addition, subtraction, etc.). 


1. Addition, and sums, of magnitudes. 

To add is προστίθημι, for the passive of which πρόσκειμαι is often 
used ; thus one segment being added ἑνὸς τμάματος ποτιτεθέντος, the 
added (straight line) ἃ ποτικειμένα, let the common HA, Ζ1 be added 
Kowal προσκείσθωσαν ai HA, ZI; the words are generally followed 
by πρός (with acc. of the thing added ἔθ), but sometimes by the 
dative, that to which the addition was made ᾧ ποτετέθη. 

For being added together we have συντίθεσθαι ; thus being added 
to itself συντιθέμενον αὐτὸ ἑαυτῷ, added together és τὸ αὐτὸ συντεθέντα, 
added to itself (continually) ἐπισυντιθέμενον ἑαυτῷ. 

Sums are commonly expressed for two magnitudes by συναμφό- 
repos used in the following different ways; the swum of BA, AA 
συναμφότερος ἡ BAA, the sum of AT, TB συναμφότερος ἡ AT, TB, the 
sum of the area and the circle τὸ συναμφότερον 6 τε κύκλος καὶ TO 
χωρίον. Again for swms in general we have such expressions as the 
line which is equal to both the radii ἡ ἴση ἀμφοτέραις ταῖς ἐκ τοῦ 
κέντρου, the line equal to (the swum of) all the lines joining ἡ ton 
πάσαις ταῖς ἐπιζευγνυούσαις. Also all the circles οἱ πάντες κύκλοι 
means the sum of all the circles; and σύγκειται ἐκ is used for is 
equal to the sum of (two other magnitudes). 

To denote plus μετά (with gen.) and σύν are. used ; together with 
the bases μετὰ τῶν βάσεων, together with half the base of the segment 
σὺν τῇ ἡμισείᾳ τῆς TOD τμήματος βάσεως ; Te and καί also express the 
same thing, and the participle of προσλαμβάνω gives another way of 
describing having something added to it; thus the squares on (all) 
the lines equal to the greatest together with the square on the greatest... 
is τὰ τετράγωνα τὰ ἀπὸ τᾶν ἰσᾶἂν τᾷ μεγίστᾳ ποτιλαμβάνοντα τό τε ἀπὸ 


τᾶς μεγίστας τετρ ἄγωνον..... 


2. Subtraction and differences. 


To subtract from is ἀφαιρεῖν ἀπό; if (the rhombus) be conceived as 
taken away ἐὰν νοηθῇ ἀφῃρημένος, let the segments be subtracted 
ἀφαιρεθέντων τὰ τμήματα. Terms common to each side in an 
equation are κοινᾶ ; the squares are common to both (sides) κοινά eVTL 
ἑκατέρων τὰ tetpaywva. Then let the common area be subtracted 
is κοινὸν ἀφῃρήσθω τὸ χωρίον, and so on; the remainder is denoted 
by the adjective λοιπός, e.g. the conical surface remaining λοιπὴ ἡ 
κωνικὴ ἐπιφάνεια. 

The difference or excess is ὑπεροχή, or more fully the eacess by 


THE TERMINOLOGY OF ARCHIMEDES. elxxvil 


which (one magnitude) exceeds (another) ὑπεροχή, ἧ ὑπερέχει... OF 
ὑπεροχά, ἃ μείζων ἐστί.... The excess is also expressed by means of 
the verb ὑπερέχειν alone ; let the difference by which the said triangles 
exceed the triangle AAT be ©, ᾧ δὴ ὑπερέχει τὰ εἰρημένα τρίγωνα τοῦ 
AAT τριγώνου ἔστω τὸ Θ, to exceed by less than the excess of the cone 
W over the half of the spheroid ὑπερέχειν ἐλάσσονι ἢ ᾧ (or ἁλίκῳ) 
ὑπερέχει ὃ Ψ κῶνος τοῦ ἡμίσεος TOD σφαιροειδέος (Where ᾧ ὑπερέχει May 
also be omitted). Again the excess may be ᾧ μείζων ἐστί. The 
opposite to ὑπερέχει is λείπεται (with gen.). 

Equal to twice a certain excess ἴσα δυσὶν ὑπεροχαῖς, with which 
equal to one excess, ἴσα μιᾷ ὑπεροχᾷ, is contrasted. 

The following sentence practically states the equivalent of an 
algebraical equation ; the rectangle under ZH, BA exceeds the rect- 
angle under ZE, EA by the (sum of) the rectangle contained by BA, 
EH and the rectangle under ZE, BE, ὑπερέχει τὸ ὑπὸ τἂν ZH, BA τοῦ 
ὑπὸ τᾶν ΖΕ), EA τῷ τε ὑπὸ τἂν ΞΔ, EH περιεχομένῳ καὶ τῷ ὑπὸ τᾶν ΖΕ, 
HE. Similarly twice PH together with ILS is (equal to) the swum of 
ΣΡ, PH, δύο μὲν ai PH μετὰ τᾶς ΠΣ συναμφότερός ἐστιν ἃ PII. 

3. Multiplication. 

To multiply is πολλαπλασιάζω; multiply one another (of numbers) 
πολλαπλασιάζειν ἀλλάλους ; to multiply by a number is expressed by 
the dative ; let A be multiplied by Θ πεπολλαπλασιάσθω 6 A τῷ ©. 

Multiplied into is sometimes ἐπί (with acc.); thus the rectangle 
HO, ΘΑ into ΘΑ (1.6. a solid figure) is τὸ ὑπὸ τῶν HO, ΘΑ ἐπὶ 
τὴν ΘΑ. 

"4, Diwision. 

To divide διαιρεῖν ; let it be divided into three equal parts at the 
points K, Θ, διῃρήσθω εἰς τρία ἴσα κατὰ τὰ K, © σαμεῖα ; to be divisible 
by μετρεῖσθαι ὑπό. 


Proportions. 


A ratio is λόγος, proportional is expressed by the phrase im 
proportion ἀνάλογον, and a proportion is ἀναλογία. We find in 
Archimedes some uses of the verb λέγω which seem to throw light 
on the definition found in Euclid of the relation or ratio between 
two magnitudes. One passage (On Conoids and Spheroids, Prop. 1) 
says if the terms similarly placed have, two and two, the same ratio 
and the first magnitudes are taken in relation to some other mag- 
nitudes in any ratios whatever εἴ κα κατὰ δύο τὸν αὐτὸν λόγον ἔχωντι 


ἘΠ. Ἃ. m 


elxxvili INTRODUCTION. 


τὰ ὁμοίως τεταγμένα, λεγήται δὲ TA πρῶτα μεγέθεα ποτί τινα ἄλλα 
μεγέθεα...ἐν λόγοις ὁποιοισοῦν, if A, B... be in relation to N, Z... but 
Z be not in relation to anything (i.e. has no term corresponding to 
it) εἴ κα... τὰ μὲν A, B,... Aeywvrar ποτὶ τὰ N, By... τὸ δὲ Z μηδὲ 
ποθ᾽ ἕν λεγήται. 

A mean proportional between is μέση ἀνάλογον τῶν..., is ὦ mean 
proportional between μέσον λόγον ἔχει THS...Kal THS..., two mean pro- 
portionals δύο μέσαι ἀνάλογον with or without κατὰ τὸ συνεχές in 
continued proportion. 

If three straight lines be proportional ἐὰν τρεῖς εὐθεῖαι ἀνάλογον 
ὦσι, a fourth proportional τετάρτα ἀνάλογον, if four straight lines be 
proportional in continued proportion εἴ κα τέσσαρες γραμμαὶ ἀνάλογον 
ἔωντι ἐν τᾷ συνεχεῖ ἀναλογίᾳ, at the ‘point dividing (the line) in the 
said proportion κατὰ τὰν ἀνάλογον τομὰν τᾷ εἰρημένᾳ. 

The ratio of one straight line to another is e.g. ὁ τῆς PA πρὸς AX 
λόγος or ὁ (λόγος), dv ἔχει ἡ PA πρὸς τὴν AX; the ratio of the bases ὃ 
τῶν βασίων λόγος ; has the ratio of 5 to 2 λόγον ἔχει, ὃν πέντε πρὸς 
δύο. 

For having the same ratio as we find the following constructions. 
Have the same ratio to one another as the bases τὸν αὐτὸν ἔχοντι λόγον 
mot ἀλλάλους ταῖς βάσεσιν, as the squares on the radi ὃν ai ἐκ τῶν 
κέντρων δυνάμει ; TA has to PZ the (linear) ratio which the square on 
TA has to the square on H, ov ἔχει λόγον ἡ TA πρὸς τὴν H δυνάμει, 
τοῦτον ἔχει Tov λόγον ἡ TA πρὸς PZ μήκει. 18 dwided wn the same 
ratio εἰς τὸν αὐτὸν λόγον τέτμηται, Or Simply ὁμοίως ; will divide the 
diameter in the proportion of the successive odd numbers, unity 
corresponding to the (part) adjacent to the vertex of the segment τὰν 
διάμετρον τεμοῦντι εἰς τοὺς τῶν ἑξῆς περισσών ἀριθμῶν λόγους, ἑνὸς 
λεγομένου ποτὶ τᾷ κορυφᾷ τοῦ τμάματος. 

To have a less (or greater) ratio than is ἔχειν λόγον ἐλάσσονα (or 
μείζονα) with the genitive of the second ratio or a phrase introduced 
by 7; to have a less ratio than the greater magnitude has to the less, 
ἔχειν λόγον ἐλάσσονα ἢ τὸ μεῖζον μέγεθος πρὸς TO ἔλασσον. 

For duplicate, triplicate etc. ratios we have the following 
expressions: has the triplicate ratio of the same ratio τριπλασίονα 
λόγον ἔχει TOD αὐτοῦ λόγου, has the duplicate ratio of EA to AK 
διπλασίονα λόγον ἔχει ἥπερ ἡ EA πρὸς AK, are in the triplicate ratio 
of the diameters in the bases ἐν τριπλασίονι λόγῳ εἰσὶ τῶν ἐν ταῖς 
βάσεσι διαμέτρων, sesquialterate ratio ἡμιόλιος λόγος. With these 
expressions must be contrasted the use of double, quadruple οἷο. 


THE TERMINOLOGY OF ARCHIMEDES. clxxix 


ratio in the sense of a simple multiple by 2, 4 etc, eg. of any 
number of areas be placed in order, each being four times the next εἴ 
κα χωρία τεθέωντι ἑξῆς ὁποσαοῦν ἐν τῷ τετραπλασίονι λόγῳ. 

The ordinary expression for a proportion is as A is to B so ws T 
to A, ws ἡ A πρὸς τὴν B, οὕτως ἡ Τὶ πρὸς τὴν A. Let AE be made so 
that AE is to TE as the sum of ΘΑ, AE ts to AE, πεποιήσθω, ὡς 
συναμφότερος ἡ ΘΑ, AE zpos τὴν AE, οὕτως ἡ AE zpos TE. The 
antecedents are τὰ ἡγούμενα, the consequents τὰ ἑπόμενα. 

For reciprocally proportional the parts of ἀντιπέπονθα are used ; 
the bases are reciprocally proportional to the heights ἀντιπεπόνθασιν 
ai βάσεις ταῖς ὕψεσιν, to be reciprocally in the same proportion 
ἀντιπεπονθέμεν κατὰ τὸν αὐτὸν λόγον. 

A ratio compounded of is λόγος συνημμένος (or συγκείμενος) ἔκ TE 
TOv...Kai Tov...; the ratio of PA to AX ἐδ equal to that compounded of 
6 τῆς PA πρὸς AX λόγος συνῆπται éx.... Two other expressions for 
compounded ratios are 6 τοῦ ἀπὸ ΑΘ -πρὸς τὸ ἀπὸ BO καὶ ὃ (or 
προσλαβὼν τὸν) τῆς ΑΘ πρὸς ΘΒ, the ratio of the square on A® to 
the square on BO multiplied by the ratio of ΑΘ to ΘΒ. 

The technical terms for transforming such a proportion as 
a:b=c:d are as follows: 


1, ἐναλλάξ alternately (usually called permutando or alternando) 
means transforming the proportion into a :c=6: d. 


2. ἀνάπαλιν reversely (usually invertendo), ὦ : ὦ -- (ὦ : 6. 


9. σύνθεσις λόγου is composition of a ratio by which the ratio 
a:b becomes a+6:6. The corresponding Greek term to com- 
ponendo is συνθέντι, which means no doubt literally “to one who 
has compounded,” i.e. “if we compound,” the ratios. Thus συνθέντι 
denotes the inference thata+6b:b=c+d:d. κατὰ σύνθεσιν is also 
used in the same sense by Archimedes. 


4. διαίρεσις λόγου signifies the division of a ratio in the sense of 
separation or subtraction by which a : ὦ becomes ὦ ---ὦ : ὦ. Similarly 
διελόντι (or κατὰ διαίρεσιν) denotes the inference that a—b:b= 
c—d:d. The translation diwidendo is therefore somewhat mis- 
leading. 


5. ἀναστροφὴ λόγου conversion of a ratio and ἀναστρέψαντι 
correspond respectively to the ratio a : ὦ -- ὦ and to the inference 
thata:a—b=c:c-—d. 


clxxx INTRODUCTION. 


6. διέ ἴσου ex aequali (sc. distantia) is applied e.g. to the 
inference from the proportions 


a:6:c:detec..=A:B:C:D etc. 
that a:a—A - ἢ. 


When this dividing-out of ratios takes place between proportions 
with corresponding terms placed crosswise, it is described as δ ἴσου 
ἐν τῇ τεταραγμένῃ ἀναλογίᾳ, ex aequali im disturbed proportion or 
ἀνομοίως τῶν λόγων τεταγμένων the ratios being dissimilarly placed ; 
this is the case e.g. when we have two proportions 


Cro=BsE eC 
Dc — ANG, 
and we infer that a:c=A:C. 


Arithmetical terms. 


Whole multiples of any magnitude are generally described as the 
double of, the triple of etc., ὃ διπλάσιος, ὃ τριπλάσιος κ-τ.λ., following 
the gender of the particular magnitude ; thus the (surface which is) 
four times the greatest circle in the sphere 4 τετραπλασία τοῦ μεγίστου 
κύκλου τῶν ἐν τῇ σφαίρᾳ ; five times the sum of AB, BE together with 
ten times the swum of ΓΒ, BA, ἃ πενταπλασία συναμφοτέρου tas AB, BE 
μετὰ τᾶς δεκαπλασίας συναμφοτέρου τᾶς TB, BA. The same multiple 
as τοσαυταπλασίων... ὁσαπλασίων ἐστί, or ἰσάκις πολλαπλασίων... καί. 
The general word for a multiple of is πολλαπλάσιος or πολλαπλασίων, 
which may be qualified by any expression denoting the number of 
times multiplied ; thus multiplied by the same number πολλαπλάσιος 
τῷ αὐτῷ ἀριθμῷ, multiples .according to the successive mumbers 
πολλαπλάσια κατὰ τοὺς ἑξῆς ἀριθμούς. 

Another method is to use the adverbial forms twice dis, thrice 
τρίς, etc., which are either followed by the nominative, e.g. twice EA 
δὶς ἡ EA, or constructed with a participle, e.g. twice taken dis Nap- 
Bavopevos or δὶς εἰρημένος ; together with twice the whole circumference 
of the circle μεθ᾽ ὅλας τᾶς τοῦ κύκλου περιφερείας Sis λαμβανομένας. 
Similarly the same number of times (the said circumference) as is 
expressed by the number one less than (that of) the revolutions 
τοσαυτάκις λαμβανομένας, ὅσος ἐστὶν ὃ ἑνὶ ἐλάσσων ἀριθμὸς τᾶν 
περιφορᾶν. An interesting phrase is the following, as many times as 
the line Τ' Δ is contained (literally added together) in AA, so many temes 
let the time ZH be contained in the time AH, ὁσάκις συγκείται ἃ TA 


THE TERMINOLOGY OF ARCHIMEDES. elxxx1 


γραμμὰ ἐν τᾷ AA, τοσαυτάκις συγκείσθω ὃ χρόνος ὃ ZH ἐν τῷ χρόνῳ τῷ 
ΛΗ. 

Submultiples are denoted by the ordinal number followed by 
μέρος ; one-seventh is ἕβδομον μέρος and so on, one-half being however 
ἥμισυς. When the denominator is a large number, a circumlocutory 
phrase is used; thus less than ;};th part of a right angle ἐλάττων ἢ 
διαιρεθείσας Tas ὀρθᾶς εἰς ρξδ΄ τούτων ἕν μέρος. 

When the numerator of a fraction is not unity, it is expressed 
by the ordinal number, and the denominator by a compound 
substantive denoting such and such a submultiple; e.g. two-thirds 
δύο τριταμόρια, three-fifths τρία πεμπταμόρια. 

There are two improper fractions which have special names, 
thus one-and-a-half of is ἡμιόλιος, one-and-a-third of ἐπίτριτος. 
Where a number is partly integral and partly fractional, the integer 
is first stated and the fraction follows introduced by καὶ ἔτι or καί 
and besides. The phrases used to express the fact that the cir- 
cumference of a circle is less than 31 but greater than 31° times its 
diameter deserve special notice ; (1) παντὸς κύκλου ἡ περίμετρος τῆς 
διαμέτρου τριπλασίων ἐστί, καὶ ἔτι ὑπερέχει ἐλάσσονι μὲν ἢ ἑβδόμῳ μέρει 
τῆς διαμέτρου, μείζονι δὲ ἢ δέκα ἑβδομηκοστομόνοις, and (2) τριπλασίων 
ἐστὶ καὶ ἐλάσσονι μὲν ἢ ἑβδόμῳ μέρει, μείζονι δὲ ἢ ( οα΄ μείζων. We 
also have the phrase for the first part ἐλάσσων ἢ τριπλασίων καὶ 
ἑβδόμῳ μέρει μείζων. 

To measure μετρεῖν, common measure κοινὸν μέτρον, commensurable, 
incommensurable σύμμετρος, ἀσύμμετρος. 


Mechanical terms. 


Mechanics τὰ μηχανικά, weight βάρος ; centre of gravity κέντρον 
tov Bapeos with another genitive of the body or magnitude; in the 
plural we have either τὰ κέντρα αὐτῶν τοῦ Bapeos or τὰ κέντρα τῶν 
βαρέων. κέντρον is also used alone. 

A lever ζυγός or ζύγιον, the horizon ὃ δρίζων ; in a vertical line is 
represented by perpendicularly κατὰ κάθετον, thus the point of 
suspension and the centre. of gravity of the body suspended are in a 
vertical line κατὰ καθετόν ἐστι TO τε σαμεῖον TOD κρεμαστοῦ καὶ TO 
κέντρον τοῦ βάρεος τοῦ κρεμαμένου. Of suspension from or αὐ ἐκ or ° 
κατά (with acc.) is used. Let the triangle be suspended from the 
points B, T, κρεμάσθω τὸ τρίγωνον ἐκ τῶν B, Τ' σαμείων ; if the 
suspension of the triangle BAT αὐ B, Τ' be set free, and it be suspended 
at B, the triangle remains in its position εἴ κα τοῦ BAT τριγώνου ἃ 


elxxxul INTRODUCTION. 


μὲν κατὰ τὰ B, Τ' κρέμασις λυθῇ, κατὰ δὲ τὸ E κρεμασθῇ, μένει τὸ 
τρίγωνον, ὡς νῦν ἔχει. 

To incline towards ῥέπειν ἐπί (800.)}; to be im equilibrium 
ἰσορροπεῖν, they will be in equilibrium with A held fast κατεχομένου 
τοῦ A ἰσορροπήσει, they will be in equilibrium at A (1.6. will balance 
about A) κατὰ τὸ A ἰσορροπησοῦντι; AB is too great to balance Τ' 
μεῖζόν ἐστι τὸ AB ἢ ὥστε ἰσορροπεῖν τῷ 1. The adjective for wm 
equilibrium is ἰσορρεπής ; let it be in equilibrium with the triangle 
TAH, ἰσορρεπὲς ἔστω τῷ TAH τριγώνῳ: To balance at certain 
distances (from the point of support or the centre of gravity of a 


system) is ἀπό τινων μακέων ἰσορροπεῖν. 


Theorems, problems, etc. 


A theorem θεώρημα (from θεωρεῖν to investigate); a problem 
πρόβλημα, with which the following expressions may be compared, 
the (questions) propounded concerning the figures τὰ προβεβλημένα 
περὶ τῶν σχημάτων, these things are propounded for investigation 
προβαλλέται τάδε θεωρήσαι ; also πρόκειμαι takes the place of the 
passive, which it was proposed (or required) to find ὅπερ προέκειτο 
εὑρεῖν. 

Another similar word is ἐπίταγμα, direction or requirement ; 
thus the theorems and directions necessary for the proofs of them τὰ 
θεωρήματα καὶ τὰ ἐπιτάγματα τὰ χρείαν ἔχοντα εἰς Tas ἀποδειξίας αὐτῶν, 
in order that the requirement may be fulfilled ὅπως γένηται τὸ ἐπι- 
ταχθέν (or ἐπίταγμα). To satisfy the requirement is ποιεῖν τὸ ἐπίταγμα 
(either e.g. of lines in a figure, or of the person solving the 
problem). 

After the setting out (ἔκθεσις) in any proposition there follows 
the short statement of what it is required to prove or to do. In 
the former case (that of a theorem) Archimedes uses one of three 
expressions δεικτέον it is required to prove, λέγω or φαμὶ δή 1 assert 
or say; and in the second case (that of a problem) δεῖ dy i ts 
required (to do so and so). 

In a problem the analysis ἀνάλυσις and synthesis σύνθεσις are 
distinguished, the latter being generally introduced with the words 
the synthesis of the problem will be as follows συντεθήσεται τὸ 
πρόβλημα οὕτως. The parts of the verb ἀναλύειν are similarly 
used ; thus the analysis and synthesis of each of these (problems) will 
be given at the end ἑκάτερα δὲ ταῦτα ἐπὶ τέλει ἀναλυθήσεταί τε καὶ 


, 
συντεθήσεται. 


THE TERMINOLOGY OF ARCHIMEDES. elxxxiil 


A notable term in connexion with problems is the διορισμός 
(determination), which means the determination of the limits within 
which a solution is possible*. If a solution is always possible, the 
problem does not involve a διορισμός, οὐκ ἔχει διορισμόν ; otherwise 
it does involve it, ἔχει διορισμόν. 


Data and hypotheses. 


For given some part of the verb δίδωμι is used, generally the 
participle δοθείς, but sometimes δεδομένος and once or twice διδόμενος. 
Let a circle be given δεδόσθω κύκλος, given two unequal magnitudes 
δύο μεγεθῶν ἀνίσων δοθέντων, each of the two lines TA, EZ is given 
ἐστὶν δοθεῖσα ἑκατέρα τῶν TA, EZ, the same ratio as the given one 
λόγος ὃ αὐτὸς τῷ δοθέντι. Similar expressions are the assigned ratio 
6 ταχθεὶς λόγος, the given area τὸ προτεθὲν (Or προκείμενον) χωρίον. 

Given in position θέσει simply (sc. δεδομένη). 

Of hypotheses the parts of the verb ὑποτίθεμαι and (for the 
passive) ὑπόκειμαι are used; with the same suppositions τῶν αὐτῶν 
ὑποκειμένων, let the said suppositions be made ὑποκείσθω τὰ εἰρημένα, 
we make these suppositions ὑποτιθέμεθα τάδε. 

Where in a reductio ad absurdum the original hypothesis is 
referred to, and generally where an earlier step is quoted, the past 
tense of the verb is used ; but it was not (so) οὐκ ἣν δέ, for it was less 
qv γὰρ ἐλάσσων, they were proved equal ἀπεδείχθησαν ἴσοι, for this has 
been proved to be possible δεδείκται yap τοῦτο δυνατὸν ἐόν. Where a 
hypothesis is thus quoted, the past tense of ὑπόκειμαι has various 
constructions after it, (1) an adjective or participle, AZ, BH were 
supposed equal ἴσαι ὑπέκειντο ai AZ, BH, it is by hypothesis a tangent 
ὑπέκειτο ἐπιψαύουσα, (2) an infinitive, for by hypothesis it does not 
cut ὑπέκειτο yap μὴ τέμνειν, the axis is by hypothesis not at right 
angies to the parallel planes ὑπέκειτο ὃ ἀξων μὴ εἶμεν ὀρθὸς ποτὶ τὰ 
παράλλαλα ἐπίπεδα, (3) the plane is supposed to have been drawn 
through the centre τὸ ἐπίπεδον ὑπόκειται διὰ τοῦ κέντρου ἄχθαι. 

Supposing it found εὑρεθέντος absolutely. Swppose it done 
γεγονέτω. 

The usual idiomatic use of εἰ δὲ μή after a negative statement 
may be mentioned ; i¢ will not meet the surface in another point, 
otherwise... ob yap ἁψέται κατ᾽ ἄλλο σαμεῖον τᾶς ἐπιφανείας: εἰ δὲ 


* Cf. Apollonius of Perga, p. Ιχχ, note, 


clxxxiv INTRODUCTION. 


Inferences, and adaptation to different cases. 


The usual equivalent for therefore is apa; οὖν and τοίνυν are 
generally used in a somewhat weaker sense to mark the starting- 
point of an argument, thus ἐπεὶ οὖν may be translated as since, then. 
Since is ἐπεί, because διότι. 

πολλῷ μᾶλλον much more then is apparently not used in Archi- 
medes, who has πολλῷ alone; thus much less then is the ratio of the 
circumscribed figure to the inscribed than that of Καὶ to Ἡ πολλῷ 
apa τὸ περιγραφὲν πρὸς τὸ ἐγγραφὲν ἐλάσσονα λόγον ἔχει τοῦ, ὃν ἔχει ἡ 
Κ πρὸς Η. 

διά with the accusative is a common way of expressing the 
reason why; because the cone is isosceles διὰ τὸ ἰσοσκελῆ εἶναι τὸν 
κῶνον, for the same reason διὰ ταὐτά. 

διά with the genitive expresses the means by which a proposition 
is proved ; by means of the construction διὰ τῆς κατασκευῆς, by the 
same means διὰ τῶν αὐτῶν, by the same method διὰ τοῦ αὐτοῦ τρόπου. 

Whenever this is the case, the surface is greater ὅταν τοῦτο ἢ, 
μείζων γίνεται ἡ ἐπιφάνεια..., if this is the case, the angle BA® is 
equal..., εἰ δὲ τοῦτο, ἴσα ἐστὶν ἃ ὑπὸ ΒΑΘ ywvia..., which is the same 
thing as showing that... 6 ταὐτόν ἐστι τῷ δεῖξαι, ὅτι... 

Similarly for the sector ὁμοίως δὲ καὶ ἐπὶ τοῦ τομέως, the proof 
is the same as (that wsed to show) that ἃ αὐτὰ ἀπόδειξις ἅπερ καὶ ὅτι, 
the proof that...is the same ἃ αὐτὰ ἀπόδειξίς ἐντι καὶ διότι..., the same 
argument holds for all rectilineal figures inscribed in the segments in 
the recognised manner (see p. 204) ἐπὶ πάντων εὐθυγράμμων τῶν 
ἐγγραφομένων ἐς τὰ τμάματα γνωρίμως ὃ αὐτὸς λόγος ; it will be possible, 
having proved it for a circle, to transfer the same argument im 
the case of the sector ἔσται ἐπὶ κύκλου δείξαντα μεταγαγεῖν τὸν ὅμοιον 
λόγον καὶ ἐπὶ τοῦ τομέως ; the rest will be the same, but τέ will be the 
lesser of the diameters which will be intercepted within the spheroid 
(instead of the greater) τὰ μὲν ἄλλα τὰ αὐτὰ ἐσσείται, τᾶν δὲ διαμέτρων 
ἃ ἐλάσσων ἐσσείται ἃ ἐναπολαφθεῖσα ἐν τῷ σφαιροειδεῖ ; it will make 


no difference whether...or...dvolcer δὲ οὐδέν, εἴτε... εἴτε.... 


Conclusions. 


The proposition is therefore obvious, or is proved δῆλον οὖν ἐστι 
(or δέδεικται) τὸ προτεθέν ; similarly φανερὸν οὖν ἐστιν, ὃ ἔδει δείξαι, 
and ἔδει δὲ τοῦτο δείξα. Which is absurd, or impossible ὅπερ ἄτοπον, 
ΟΥ ἀδύνατον. 

A curious use of two negatives is contained in the following: 


THE TERMINOLOGY OF ARCHIMEDES. elxxxv 


οὐκ apa οὔκ ἐστι κέντρον τοῦ Bapeos τοῦ ΔΕΖ τριγώνου τὸ N σαμεῖον. 
ἔστιν apa, therefore it is not possible that the point N should not be the 
centre of gravity of the triangle AEZ. It must therefore be so. 

Thus a rhombus will have been formed ἔσται δὴ γεγονὼς ῥόμβος ; 
two unequal straight lines have been found satisfying the requirement 


ὑ Ξ ἰσὶν ἄρα δύο εὐθεῖαι ἃ οιοῦσαι τὸ ἐπίταγμα 
εὑρημέναι εἰσὶν apa δύο εὐθεῖαι ἀνισοι 7 α γμα. 


Direction, concavity, convexity. 


In the same direction ἐπὶ τὰ αὐτά, in the other direction ἐπὶ 
τὰ ἕτερα, concave in the same direction ἐπὶ τὰ αὐτὰ κοίλη ; in the same 
direction as ἐπὶ τὰ αὐτά with the dative or ἐφ᾽ a, thus in the same 
direction as the vertex of the cone ἐπὶ τὰ αὐτὰ τᾷ TOD κώνου κορυφᾷ, 
drawn in the same direction as (that of) the convex side of it ἐπὶ τὰ 
αὐτὰ ἀγομέναι, ἐφ᾽ a ἐντι τὰ κυρτὰ αὐτοῦ. For on the same side of ἐπὶ 
τὰ αὐτά is followed by the genitive, they fall on the same side of the 
line ἐπὶ τὰ αὐτὰ πίπτουσι τῆς γραμμῆς. 

On each side of ἐφ᾽ ἑκάτερα (with gen.); on each side of the plane 
of the base ἐφ᾽ ἑκάτερα τοῦ ἐπιπέδου τῆς βάσεως. 


Miscellaneous. 


Property σύμπτωμα. Proceeding thus continually, ἀεὶ τοῦτο 
ποιοῦντες, ἀεὶ τούτου γενομένου, OY τούτου ἑξῆς γινομένου. In the 
elements ἐν τῇ στοιχειώσει. 

One special difference between our terminology and the Greek is 
that whereas we speak of any circle, any straight line and the like, 
the Greeks say every circle, every straight line, etc. Thus any 
pyramid is one third part of the prism with the same base as the 
pyramid and equal height πᾶσα πυραμὶς τρίτον μέρος ἐστὶ τοῦ πρίσματος 
τοῦ τὰν αὐτὰν βάσιν ἔχοντος τᾷ πυραμίδι καὶ ὕψος ἴσον. 7 define the 
diameter of any segment as διάμετρον καλέω παντὸς τμάματος. 70 
exceed any assigned (magnitude) of those which are comparable with 
one another ὑπερέχειν παντὸς τοῦ προτεθέντος τῶν πρὸς ἄλληλα 
λεγομένων. 

Another noteworthy difference is illustrated in the last sentence. 
The Greeks did not speak as we do of @ given area, ὦ given ratio 
etc., but of the given area, the given ratio, and the like. Thus J/¢ is 
possible...to leave certain segments less than a given area δυνατόν 
ἐστιν... λείπειν τινα τμήματα, ἅπερ ἔσται ἐλάσσονα τοῦ προκειμένου 
χωρίου ; to divide a given sphere by a plane so that the segments have 
to one another an assigned ratio τὰν δοθεῖσαν σφαῖραν ἐπιπέδῳ τεμεῖν, 


7 lol 
ὥστε TA τμάματα αὐτᾶς ToT ἀλλαλα τὸν ταχθέντα λόγον ἔχειν. 


H. A. γ7ὺ 


elxxxvl INTRODUCTION. 


Magnitudes in arithmetical progression are said to exceed each 
other by an equal (amount) ; of there be any number of magnitudes in 
arithmetical progression εἴ κα ἔωντι μεγέθεα ὁποσαοῦν τῷ ἴσῳ ἀλλάλων 
ὑπερέχοντα. The common difference is the eacess ὑπεροχά, and the 
terms collectively are spoken of as the magnitudes exceeding by the 
equal (difference) τὰ τῷ ἴσῳ ὑπερέχοντα. The least term is τὸ ἐλάχιστον, 
the greatest term τὸ μέγιστον. The swm of the terms is expressed by 
πάντα τὰ τῷ low ὑπερέχοντα. ; 

Terms of a geometrical progression are simply im (continued) 
proportion ἀνάλογον, the series is then 9 ἀναλογία, the proportion, 
and a term of the series is τὶς τῶν ἐν τᾷ αὐτᾷ avadoyia. Numbers in 
geometrical progression beginning from unity are ἀριθμοὶ ἀνάλογον 
ἀπὸ μονάδος. Let the term A of the progression be taken which 
is distant the same number of terms from © as A is distant from 
unity λελάφθω ἐκ τᾶς ἀναλογίας ὁ A ἀπέχων ἀπὸ τοῦ Θ τοσούτους, ὅσους 


ε > Ν Ψ' > , 
6 A ἀπὸ μονάδος ἀπέχει. 


THE WORKS OF 


ARCHIMEDES. 


ON THE SPHERE AND CYLINDER. 
BOOK I. 


‘“ ARCHIMEDES to Dositheus greeting. 


On a former occasion I sent you the investigations which 
I had up to that time completed, including the proofs, showing 
that any segment bounded by a straight line and a section of a 
right-angled cone [a parabola] is four-thirds of the triangle 
which has the same base with the segment and equal height. 
Since then certain theorems not hitherto demonstrated (ave- 
λέγκτων) have occurred to me, and I have worked out the proofs 
of them. They are these: first, that the surface of any sphere 
is four times its greatest circle (τοῦ μεγίστου κύκλου); next, 
that the surface of any segment of a sphere is equal to a circle 
whose radius (ἡ ἐκ τοῦ κέντρου) is equal to the straight line 
drawn from the vertex (κορυφή) of the segment to the circum- 
ference of the circle which is the base of the segment; and, 
further, that any cylinder having its base equal to the greatest 
circle of those in the sphere, and height equal to the diameter 
of the sphere, is itself [1.6. in content] half as large again as the 
sphere, and its surface also [including its bases] is half as large 
again as the surface of the sphere. Now these properties were 
all along naturally inherent in the figures referred to (αὐτῇ τῇ 
φύσει προυπῆρχεν περὶ τὰ εἰρημένα σχήματα), but remained 
unknown to those who were before my time engaged in the 
study of geometry. Having, however, now discovered that the 
properties are true of these figures, I cannot feel any hesitation 

H. A, 1 


2 ARCHIMEDES 


in setting them side by side both with my former investiga- 
tions and with those of the theorems of Eudoxus on solids 
which are held to be most irrefragably established, namely, 
that any pyramid is one third part of the prism which has the 
same base with the pyramid and equal height, and that any 
cone is one third part of the cylinder which has the same 
base with the cone and equal height. For, though these 
properties also were naturally inherent in the figures all along, 
yet they were in fact unknown to all the many able geometers 
who lived before Eudoxus, and had not been observed by any 
one. Now, however, it will be open to those who possess the 
requisite ability to examine these discoveries of mine. They 
ought to have been published while Conon was still alive, 
for I should conceive that he would best have been able to 
grasp them and to pronounce upon them the appropriate 
verdict ; but, as I judge it well to communicate them to those 
who are conversant with mathematics, I send them to you with 
the proofs written out, which it will be open to mathematicians 
to examine. Farewell. 


I first set out the axioms* and the assumptions which I 
have used for the proofs of my propositions. 


DEFINITIONS. 


1. There are in a plane certain terminated bent lines 
(καμπύλαι γραμμαὶ πεπερασμέναι), which either lie wholly on 
the same side of the straight lines joining their extremities, or 
have no part of them on the other side. 


2. I apply the term concave in the same direction 
to a line such that, if any two points on it are taken, either 
all the straight lines connecting the points fall on the same 
side of the line, or some fall on one and the same side while 
others fall on the line itself, but none on the other side. 


* Though the word used is ἀξιώματα, the “axioms” are more of the nature 
of definitions ; and in fact Eutocius in his notes speaks of them as such (ὅροι). 

+ Under the term bent line Archimedes includes not only curved lines of 
continuous curvature, but lines made up of any number of lines which may be 
either straight or curved. 


ON THE SPHERE AND CYLINDER I. 3 


3. Similarly also there are certain terminated surfaces, not 
themselves being in a plane but having their extremities in a 
plane, and such that they will either be wholly on the same 
side of the plane containing their extremities, or have no part 
of them on the other side. 


4. I apply the term concave in the same direction 
to surfaces such that, if any two points on them are taken, the 
straight lines connecting the points either all fall on the same 
side of the surface, or some fall on one and the same side of 
it while some fall upon it, but none on the other side. 


5. I use the term solid sector, when a cone cuts a sphere, 
and has its apex at the centre of the sphere, to denote the 
figure comprehended by the surface of the cone and the surface 
of the sphere included within the cone. 


6. I apply the term solid rhombus, when two cones with 
the same base have their apices on opposite sides of the plane 
of the base in such a position that their axes lie in a straight 
line, to denote the solid figure made up of both the cones. 


ASSUMPTIONS. 


1. Of all lines which have the same extremities the straight 
line is the least*. 


* This well-known Archimedean assumption is scarcely, as it stands, a 
definition of a straight line, though Proclus says [p. 110 ed. Friedlein] ‘‘ Archi- 
medes defined (ὡρίσατο) the straight line as the least of those [lines] which have 
the same extremities. For because, as Euclid’s definition says, ἐξ ἴσου κεῖται τοῖς 
ἐφ᾽ ἑαυτῆς σημείοις, it is in consequence the least of those which have the same 
extremities.” Proclus had just before [p. 109] explained Euclid’s definition, 
which, as will be seen, is different from the ordinary version given in our text- 
books; a straight line is not ‘‘that which lies evenly between its extreme points,” 
but ‘‘that which ἐξ ἴσου τοῖς ἐφ᾽ ἑαυτῆς σημείοις κεῖται." The words of Proclus 
are, ‘‘ He [Euclid] shows by means of this that the straight line alone [of all 
lines] occupies a distance (κατέχειν διάστημα) equal to that between the points 
onit. For, as far as one of its points is removed from another, so great is the 
length (μέγεθος) of the straight line of which the points are the extremities ; 
and this is the meaning of τὸ ἐξ ἴσου κεῖσθαι τοῖς ἐφ᾽ ἑαυτῆς σημείοις. But, if you 
take two points on a circumference or any other line, the distance cut off 
between them along the line is greater than the interval separating them; and 
this is the case with every line except the straight line.’ It appears then from 
this that Euclid’s definition should be understood in a sense very like that of 


1—2 


4 i ARCHIMEDES 


2. Of other lines in a plane and having the same extremi- 
ties, [any two] such are unequal whenever both are concave in 
the same direction and one of them is either wholly included 
between the other and the straight line which has the same 
extremities with it, or is partly included by, and is partly 
common with, the other; and that [line] which is included is 
the lesser [of the two]. 


3. Similarly, of surfaces which have the same extremities, 
if those extremities are in a plane, the plane is the least [in 
area]. 

4. Of other surfaces with the same extremities, the ex- 
tremities being in a plane, [any two] such are unequal when- 
ever both are concave in the same direction and one surface 
is either wholly included between the other and the plane which - 
has the same extremities with it, or is partly included by, and 
partly common with, the other; and that [surface] which is 
included is the lesser [of the two in area]. 


5. Further, of unequal lines, unequal surfaces, and unequal 
solids, the greater exceeds the less by such a magnitude as, 
when added to itself, can be made to exceed any assigned 
magnitude among those which are comparable with [it and 
with] one another*. 


These things being premised, ἐγ a polygon be inscribed in a 
circle, it is plain that the perimeter of the inscribed polygon is 
less than the circumference of the circle; for each of the sides 
of the polygon is less than that part of the circumference of the 
circle which is cut off by it.” 


Archimedes’ assumption, and we might perhaps translate as follows, ‘A straight 
line is that which extends equally (ἐξ ἔσου κεῖται) with the points on it,” or, to 
follow Proclus’ interpretation more closely, ‘‘A straight line is that which 
represents equal extension with [the distances separating] the points on it.” 

* With regard to this assumption compare the Introduction, chapter 11. ὃ 2. 


Or 


ON THE SPHERE AND CYLINDER I. 


Proposition 1. 


If a polygon be circumscribed about a circle, the perimeter 
of the circumscribed polygon is greater A 
than the perimeter of the circle. 8 

Let any two adjacent sides, meet- 
ing in A, touch the circle at P, Q 
respectively. 

Then [Assumptions, 2] 

PA+AQ>(are PQ). 

A similar imequality holds for each 
angle of the polygon; and, by ad- 
dition, the required result follows. 


Proposition 2. 


Given two unequal magnitudes, it is possible to find two un- 
equal straight lines such that the greater straight line has to the 
less a ratio less than the greater magnitude has to the less. 


Let AB, D represent the two unequal magnitudes, AB being 
the greater. 

Suppose BC measured along BA equal to D, and let GH be 
any straight line. : 

Then, if CA be added to itself a sufficient A 
number of times, the sum will exceed D. Let ἢ 


AF be this sum, and take # on @H produced 


such that GH is the same multiple of HZ that 7 
AF is of AC. 
Thus BA HG — AC » AF. Ὁ 
But, since AF’ > D (or CB), 
AC: AF<AC: CB. Β 
Therefore, componendo, G 


EG:GH<AB: D. 
Hence EG, GH are two lines satisfying the given condition. 


6 ARCHIMEDES 


Proposition 3. 


Given two unequal magnitudes and a circle, it is possible to 
inscribe a polygon in the circle and to describe another about it 
so that the side of the circumscribed polygon may have to the side 
of the inscribed polygon a ratio less than that of the greater 
magnitude to the less. 


Let A, B represent the given magnitudes, A being the 


greater. 


Find [Prop. 2] two straight lines F, KL, of which F is the 
greater, such that 


BSNS . 


L M 


Draw LM perpendicular to LK and of such length that 
KM=F. 


In the given circle let CH, DG be two diameters at right 
angles. Then, bisecting the angle DOC, bisecting the half 
again, and so on, we shall arrive ultimately at an angle (as 


NOC) less than twice the angle LKM. 


Join VC, which (by the construction) will be the side of a 
regular polygon inscribed in the circle. Let OP be the radius 
of the circle bisecting the angle NOC (and therefore bisecting 
NC at right angles, in H, say), and let the tangent at P meet 
OC, ON produced in S, 7’ respectively. 


Now, since ZCON ΞΖ ΕΙΣ 
ZHOC <2 LKM, 


ON THE SPHERE AND CYLINDER I. rf 


and the angles at H, L are right ; 
therefore MK: LK >OC: OH 
SOP = OF. 
Hence Se CN a Mis LK 
ἜΝ: ΚΕ: 
therefore, a fortiori, by (1), 
po CN <A Bs, 
Thus two polygons are found satisfying the given condition. 


Proposition 4. 


Again, given two unequal magnitudes and ὦ sector, it 18 
possible to describe a polygon about the sector and to inscribe 
another in it so that the side of the circumscribed polygon may 
have to the side of the inscribed polygon a ratio less than the 
greater magnitude has to the less. 


[The “inscribed polygon” found in this proposition is one 
which has for two sides the two radii bounding the sector, while 
the remaining sides (the number of which is, by construction, 
some power of 2) subtend equal parts of the are of the sector ; 
the “circumscribed polygon” is formed by the tangents parallel 
to the sides of the inscribed polygon and by the two bounding 
radu produced. ] 


In this case we make the same construction as in the last 
proposition except that we bisect the angle COD of the sector, 
instead of the right angle between two diameters, then bisect 
the half again, and so on. The proof is exactly similar to the 
preceding one. 


8 ARCHIMEDES 


Proposition 5. 


Given a circle and two unequal magnitudes, to describe a 
polygon about the circle and inscribe another in it, so that the 
circumscribed polygon may have to the inscribed a ratio less than 
the greater magmtude has to the less. 


Let A be the given circle and B, C the given magnitudes, B 
being the greater. 


F 


Take two unequal straight lines D, H, of which D is the 
greater, such that D: #<B:C [Prop. 2], and let # be a mean 
proportional between D, Μ΄, so that D is also greater than F. 


Describe (in the manner of Prop. 3) one polygon about the 
circle, and inscribe another in it, so that the side of the former 
has to the side of the latter a ratio less than the ratio D : F. 


Thus the duplicate ratio of the side of the former polygon 
to the side of the latter is less than the ratio D? : δ᾽" 


But the said duplicate ratio of the sides is equal to the 
ratio of the areas of the polygons, since they are similar ; 


therefore the area of the circumscribed polygon has to the 
area of the inscribed polygon a ratio less than the ratio D® : F”, 
or D: FE, and a fortiori less than the ratio B: C. 


ON THE SPHERE AND CYLINDER I. 9 


Proposition 6. 


“Similarly we can show that, given two unequal magnitudes 
and a sector, it ts possible to circumscribe a polygon about the 
sector and inscribe in it another similar one so that the circum- 
scribed may have to the inscribed a ratio less than the greater 
magnitude has to the less. 


And it is likewise clear that, if a circle or a sector, as well 
as ὦ certain area, be given, it is possible, by inscribing regular 
polygons in the circle or sector, and by continually imscribing 
such in the remaining segments, to leave segments of the circle or 
sector which are [together] less than the given area. For this is 
proved in the Hlements [Eucl. x11. 2]. 


But it is yet to be proved that, given a circle or sector and 
an area, it is possible to describe a polygon about the circle or 
sector, such that the area remaining between the circumference 
and the circumscribed figure is less than the given area.” 


CSE 


The proof for the circle (which, as Archimedes says, can be 
equally applied to a sector) is as follows. 


Let A be the given circle and B the given area. 


Now, there being two unequal magnitudes A + B and A, let 
a polygon (C) be circumscribed about the circle and a polygon 
(Z) inscribed in it [as in Prop. 5], so that 


eed a LS a ti hae ee τος, (1). 
The circumscribed polygon (C) shall be that required. 


10 ARCHIMEDES 


For the circle (A) is greater than the inscribed polygon (JZ). 
Therefore, from (1), a fortiori, 
C:A<A+B:A, 
whence C<A+B8, 
or C-A<B. 


Proposition 7. 


If in an isosceles cone [1.6. a right circular cone] a pyramid 
be inscribed having an equilateral base, the surface of the 
pyramid excluding the base is equal to a triangle having tts 
base equal to the perimeter of the base of the pyramid and tts 
height equal to the perpendicular drawn from the apex on one 
side of the base. 


Since the sides of the base of the pyramid are equal, it 
follows that the perpendiculars from the apex to all the sides 
of the base are equal; and the proof of the proposition is 
obvious. 


Proposition 8. 


If a pyramid be circumscribed about an isosceles cone, the 
surface of the pyramid excluding its base is equal to a triangle 
having its base equal to the perimeter of the base of the pyramid 
and its height equal to the side [1.e. a generator] of the cone. 


The base of the pyramid is a polygon circumscribed about 
the circular base of the cone, and the line joining the apex of 
the cone or pyramid to the point of contact of any side of the 
polygon is perpendicular to that side. Also all these perpen- 
diculars, being generators of the cone, are equal; whence the 
proposition follows immediately. 


ON THE SPHERE AND CYLINDER 1. 11 


Proposition 9. 


Lf in the circular base of an isosceles cone a chord be placed, 
and from its extrenuties straight lines be drawn to the apex of 
the cone, the triangle so formed will be less than the portion of 
the surface of the cone intercepted between the lines drawn to the 
apex. 

Let ABC be the circular base of the cone, and O its apex. 


Draw a chord AB in the circle, and join OA, OB. Bisect 
the are ACB in C, and join AC, BC, OC. 


Then A OAC +A OBC> A OAB. 


Let the excess of the sum of the first two triangles over the 
third be equal to the area D. 


Then JD is either less than the sum of the segments 4 EC, 
CFB, or not less. 


I. Let D be not less than the sum of the segments referred 
to. 

We have now two surfaces 

(1) that consisting of the portion OAEC of the surface 
of the cone together with the segment AZ#C, and 

(2) the triangle OAC; 
and, since the two surfaces have the same extremities (the 


perimeter of the triangle OAC), the former surface is greater 
than the latter, which is included by it [Asswmptions, 3 or 4]. 


12 ARCHIMEDES 


Hence (surface OA EC)+ (segment AEC) >A OAC. 
Similarly (surface OCFB) + (segment CFB) >A OBC. 


Therefore, since D is not less than the sum of the segments, 
we have, by addition, 


(surface OAHCFB)+D>AOAC+ AOBC 
>A OAB+ D, by hypothesis. 
Taking away the common part D, we have the required 
result. 
II. Let D be less than the sum of the segments AEC, 
CFB. 


If now we bisect the arcs AC, CB, then bisect the halves, 
and so on, we shall ultimately leave segments which are 
together less than D. [Prop. 6] 


Let AGH, EHC, CKF, FLB be those segments, and join 
OF, OF. 


Then, as before, 

(surface OAGE) + (segment AGE) > A OAH 
and (surface OF HC) + (segment LHC) > A OKC. 
Therefore (surface OAGHC) + (segments AGH, EHC) 

>AOAE+ AOEC 
> AOAC, a fortiori. 


Similarly for the part of the surface of the cone bounded by 
OC, OB and the are CFB. 


Hence, by addition, 
(surface OAGHHCKFLB)+ (segments AGE, EHC, CKF, FLB) 
>AOAC+ AOBC 
> AOAB + D, by hypothesis. 


But the sum of the segments is less than D, and the re- 
quired result follows. 


ON THE SPHERE AND CYLINDER I. 13 


Proposition 10. 


If in the plane of the circular base of an isosceles cone two 
tangents be drawn to the circle meeting in a point, and the points 
of contact and the point of concourse of the tangents be respectively 
joined to the apex of the cone, the sum of the two triangles 
formed by the joining lines and the two tangents are together 
greater than the included portion of the surface of the cone. 


Let ABC be the circular base of the cone, O its apex, AD, 
BD the two tangents to the circle meeting in D. Join OA, 
OB, OD. 

Let HCF be drawn touching the circle at C, the middle 
point of the arc ACB, and therefore parallel to AB. Join 
OE, OF. 


Then ED+DF> EF, 
and, adding AH + FB to each side, 
AD+ DB>AE+ EF + FB. 


Now OA, OC, OB, being generators of the cone, are equal, 
and they are respectively perpendicular to the tangents at A, 
Ὁ B. 


14 ARCHIMEDES 


It follows that 
AOAD+ AODB>AOAE+A OEF+A OFB. 


Let the area G be equal to the excess of the first sum over 
the second. 


G is then either less, or not less, than the sum of the spaces 
EAHC, FCKB remaining between the circle and the tangents, 
which sum we will call L. 


I. Let G be not less than Z. 
We have now two surfaces 


(1) that of the pyramid with apex O and base 4.0.8, 
excluding the face OAB, 


(2) that consisting of the part OACB of the surface of the 
cone together with the segment ACB. 


These two surfaces have the same extremities, viz. the 
perimeter of the triangle OAB, and, since the former includes 
the latter, the former is the greater [Assumptions, 4]. 


That is, the surface of the pyramid exclusive of the face 
OAB is greater than the sum of the surface OACB and the 
segment ACB. 


Taking away the segment from each sum, we have 

A OAE+A OEFF+A OFB +I >the surface OAHCKB. 

And G is not less than L. 

It follows that 

AOAE+A OEF +A OFB+G, 

which is by hypothesis equal to AOAD+A ODB, is greater 
than the same surface. 

II. Let G be less than L. 


If we bisect the ares AC, CB and draw tangents at their 
middle points, then bisect the halves and draw tangents, and 
so on, we shall lastly arrive at a polygon such that the sum 
of the parts remaining between the sides of the polygon and 
the circumference of the segment is less than G. 


ON THE SPHERE AND CYLINDER I. 15 


Let the remainders be those between the segment and the 
polygon APQRSB, and let their sum be M. Join OP WOQ, 
ete. 

Then, as before, 

AOAE+A OFF+A OFB>AOAP+A0PQ+...+A OSB. 

Also, as before, 

(surface of pyramid OAPQRSB excluding the face OAB) 

>the part OACB of the surface of the 
cone together with the segment ACB. 

Taking away the segment from each sum, 

A OAP+AO0PQ+...+M> the part OACB of the 
surface of the cone. 

Hence, a fortrore, 

AOAE+ A OEF+ A OFB+G, 
which is by hypothesis equal to 
AOAD +A ODB, 
is greater than the part OACB of the surface of the cone. 


Proposition 11. 


If a plane parallel to the axis of a right cylinder cut the 
cylinder, the part of the surface of the cylinder cut off by the 
plane ws greater than the area of the parallelogram in which the 
plane cuts τί. 


Proposition 12. 


If at the extremities of two generators of any right cylinder 
tangents be drawn to the circular bases in the planes of those 
bases respectively, and if the pairs of tangents meet, the 
parallelograms formed by each generator and the two corre- 
sponding tangents respectively are together greater than the 


imeluded portion of the surface of the cylinder between the two 
generators. 


[The proofs of these two propositions follow exactly the 
methods of Props. 9, 10 respectively, and it is therefore un- 
necessary to reproduce them.] 


16 ARCHIMEDES 


“From the properties thus proved it is clear (1) that, 7f a 
pyramid be inscribed in an isosceles cone, the surface of the 
pyramid excluding the base is less than the surface of the cone 
[excluding the base], and (2) that, if a pyramid be circumscribed 
about an isosceles cone, the surface of the pyramid excluding the 
base is greater than the surface of the cone eacluding the base. 


“Tt is also clear from what has been proved both (1) that, 
if a prism be inscribed in a right cylinder, the surface of the 
prism made up of its parallelograms [i.e. excluding its bases] 2s 
less than the surface of the cylinder excluding its bases, and 
(2) that, if a prism be circumscribed about a right cylinder, the 
surface of the prism made up of its parallelograms is greater 
than the surface of the cylinder excluding its bases.” 


Proposition 13. 


The surface of any right cylinder excluding the bases 15 equal 
to a circle whose radius is a mean proportional between the side 
[1.e. a generator] of the cylinder and the diameter of its base. 


Let the base of the cylinder be the circle A, and make CD 
equal to the diameter of this circle, and #F equal to the height 
of the cylinder. 


ON THE SPHERE AND CYLINDER I. ΠῚ 


Let H be a mean proportional between CD, EF, and B 
a circle with radius equal to H. 


Then the circle B shall be equal to the surface of the 
cylinder (excluding the bases), which we will call S. 


For, if not, B must be either greater or less than S. 


I. Suppose B< S. 

Then it is possible to circumscribe a regular polygon about 
B, and to inscribe another in it, such that the ratio of the 
former to the latter is less than the ratio S : B. 

Suppose this done, and circumscribe about A a polygon 
similar to that described about B; then erect on the polygon 
about A a prism of the same height as the cylinder. The 
prism will therefore be circumscribed to the cylinder, 


Let KD, perpendicular to CD, and FL, perpendicular to 
EF, be each equal to the perimeter of the polygon about A. 
Bisect CD in M, and join MK. 


Then A KDM =the polygon about A. 
Also 67 Ei.= surface of prism (excluding bases). 
Produce FE to N so that FE = EN, and join NL. 


Now the polygons about A, B, being similar, are in the 
duplicate ratio of the radii of A, B. 


Thus 

A KDM : (polygon about B) = MD? : H* 
Ξε MIP CD. bE 
=D ONE 


=AKDM: ALFN 


(since DK = FT). 
Therefore (polygon about B)=A LFN 


Ξ- EL 
= (surface of prism about A), 


from above. 


But (polygon about B) : (polygon in B)< S: B. 


ἘΠ. A; 


bo 


18 ARCHIMEDES 


Therefore 
(surface of prism about A): (polygon in B)<S : B, 
and, alternately, 
(surface of prism about A): S<(polygon in B): B; 
which is impossible, since the surface of the prism is greater 
than S, while the polygon inscribed in B is less than B. 

Therefore Be€s. 

II. Suppose B>S. 

Let a regular polygon be circumscribed about B and another 
inscribed in it so that 

(polygon about B): (polygon in B)< δ: 5. 

Inscribe in A a polygon similar to that inscribed in B, and 
erect a prism on the polygon inscribed in A of the same height 
as the cylinder. 

Again, let DK, FL, drawn as before, be each equal to the 
perimeter of the polygon inscribed in A. 

Then, in this case, 

A KDM > (polygon inscribed in A) 
(since the perpendicular from the centre on a side of the 
polygon is less than the radius of A). 

Also ALFN = 67 EL =surface of prism (excluding bases). 

Now 
(polygon in A) : (polygon in B)= MD* : H’, 

= AKDM : ALFN, as before. 
And AKDM > (polygon in A). 
Therefore 
A LFN, or (surface of prism) > (polygon in B). 
But this is impossible, because 
(polygon about B) : (polygon in B)< Β : 5, 
< (polygon about B): S, a fortiors, 
so that (polygon in B) >S8, 
> (surface of prism), a fortiort. 
Hence B is neither greater nor less than S, and therefore 
Bs: 


ON THE SPHERE AND CYLINDER I. 19 


Proposition 14. 


The surface of any isosceles cone excluding the base is equal 
to a circle whose radius is a mean proportional between the side 
of the cone [a generator] and the radius of the circle which is the 
base of the cone. 


Let the circle A be the base of the cone; draw C equal to 
the radius of the circle, and D equal to the side of the cone, and 
let H be a mean proportional between C, D. 


D 


Draw a circle B with radius equal to £. 


Then shall B be equal to the surface of the cone (excluding 
the base), which we will call S. 


If not, B must be either greater or less than ΑΚ. 


I. Suppose B< S. 


Let a regular polygon be described about B and a similar 
one inscribed in it such that the former has to the latter a ratio 
less than the ratio S : B. 


Describe about A another similar polygon, and on it set up 
a pyramid with apex the same as that of the cone. 


Then (polygon about A): (polygon about B) 
ΞΞ oe i 
5) 
= (polygon about A) : (surface of pyramid excluding base). 


2—2 


20 ARCHIMEDES 


Therefore 
(surface of pyramid) = (polygon about 8). 
Now (polygon about B) : (polygon in 8) « 8S: 8. 
Therefore 
(surface of pyramid) : (polygon in B)<S: B, 
which is impossible, (because the surface of the pyramid is 
greater than S, while the polygon in B is less than 5B). 
Hence B¢€s. 


II. Suppose B>S. 


Take regular polygons circumscribed and inscribed to B such 
that the ratio of the former to the latter is less than the ratio 
Bas. 

Inscribe in A a similar polygon to that inscribed in B, and 
erect a pyramid on the polygon inscribed in A with apex the 
same as that of the cone. 


In this case 
(polygon in A) : (polygon in B) = C? : E” 
=D 
> (polygon in A) : (surface of pyramid excluding base). 


This is clear because the ratio of C to D is greater than the 
ratio of the perpendicular from the centre of A on a side of the 
polygon to the perpendicular from the apex of the cone on the 
same side*, 


Therefore 
(surface of pyramid) > (polygon in B). 
But (polygon about B) : (polygon in b)< B: 8. 
Therefore, a fortiort, 
(polygon about B) : (surface of pyramid)< B: 5; 
which is impossible. 
Since therefore B is neither greater nor less than S, 


ip: 


* This is of course the geometrical equivalent of saying that, if a, B be two 
angles each less than a right angle, and a>, then sin a>sin β. 


ON THE SPHERE AND CYLINDER I. 21 


Proposition 15. 


The surface of any isosceles cone has the same ratio to tts 
base as the side of the cone has to the radius of the base. 


By Prop. 14, the surface of the cone is equal to a circle 
whose radius is a mean proportional between the side of the 
cone and the radius of the base. 


Hence, since circles are to one another as the squares of 
their radii, the proposition follows. 


Proposition 16. 


If an isosceles cone be cut by a plane parallel to the base, the 
portion of the surface of the cone between the parallel planes is 
equal to a circle whose radius is a mean proportional between (1) 
the portion of the side of the cone intercepted by the parallel 
planes and (2) the line which is equal to the sum of the radii of 
the circles in the parallel planes. 


Let OAB be a triangle through the axis of a cone, DE£ its 
intersection with the plane cutting off the 
frustum, and OFC the axis of the cone. re 
Then the surface of the cone OAB is 
equal to a circle whose radius is equal to 
VOA.AC. [Prop. 14.] 
Similarly the surface of the cone ODE = Ε 
is equal to a circle whose radius is equal 
to VOD. DF. 
And the surface of the frustum is 
equal to the difference between the two circles. 


A Cc B 


Now 
OA.AC—OD.DF=DA.AC+0D.AC— OD. DF. 
But OD.AC=O0A.DF, 


since OA: AC=OD: DF. 


22 ARCHIMEDES 


Hence OA.AC—OD.DF=DA.AC+DA.DF 
=DA.(AC+ DF). 
And, since circles are to one another as the squares of their 
radii, it follows that the difference between the circles whose 
radii are VOA.AC, VOD. DF respectively is equal to a circle 
whose radius is VDA .(AC + DF). 


Therefore the surface of the frustum is equal to this circle. 


Lemmas. 


“1, Cones having equal height have the same ratio as their 
bases; and those having equal bases have the same ratio as their 
heights*. 


2. If a cylinder be cut by a plane parallel to the base, then, 
as the cylinder is to the cylinder, so is the axis to the axis Ὕ. 


. 


3. The cones which have the same bases as the cylinders [and 
equal height| are in the same ratio as the cylinders. 


4. Also the bases of equal cones are reciprocally proportional 
to their heights; and those cones whose bases are reciprocally 
proportional to their heights are equal }. 


5. Also the cones, the diameters of whose bases have the same 
ratio as their axes, are to one another in the triplicate ratio of the 
diameters of the bases. 


And all these propositions have been proved by earlier 
geometers.” 


* Huclid x1r, 11. ‘Cones and cylinders of equal height are to one another 
as their bases.” 

Euclid x1. 14. ‘Cones and cylinders on equal bases are to one another as 
their heights.” 

+ Euclid x11. 13. ‘If a cylinder be cut by a plane parallel to the opposite 
planes [the bases], then, as the cylinder is to the cylinder, so will the axis be 
to the axis.” 

+ Euclid x11. 15. ‘‘ The bases of equal cones and cylinders are reciprocally 
proportional to their heights; and those cones and cylinders whose bases are 
reciprocally proportional to their heights are equal.” 

§ Euclid x11. 12. ‘Similar cones and cylinders are to one another in the 
triplicate ratio of the diameters of their bases.” 


ON THE SPHERE AND CYLINDER I. 23 


Proposition 17. 


If there be two isosceles cones, and the surface of one cone be 
equal to the base of the other, while the perpendicular from the 
centre of the base [of the first cone] on the side of that cone ws 
equal to the height [of the second], the cones will be equal. 


Let OAB, DEF be triangles through the axes of two cones 
respectively, C, G the centres of the respective bases, GH the 


A 


H 


B 


perpendicular from G on F'D; and suppose that the base of the 
cone OAB is equal to the surface of the cone DEF, and 
that OC = GH. 


Then, since the base of OAB is equal to the surface of 
DEF, 


(base of cone OAB) : (base of cone DEF) 
=(surface of DHF) : (base of DEF’) 


=DP* FG [Prop. 15] 
= DG: GH, by similar triangles, 
= DG: OC. 


Therefore the bases of the cones are reciprocally propor- 
tional to their heights; whence the cones are equal. [Lemma 


24, ARCHIMEDES 


Proposition 18. 


Any solid rhombus consisting of isosceles cones is equal to 
the cone which has its base equal to the surface of one of the 
cones composing the rhombus and its height equal to the perpen- 
dicular drawn from the apex of the second cone to one side of 
the first cone. 

Let the rhombus be OABD consisting of two cones with 
apices O, D and with a common base (the circle about AB as 
diameter). 


H G K 


D M 
V 
E 
N 


Let FHK be another cone with base equal to the surface of 
the cone OAB and height FG equal to DE, the perpendicular 
from D on OB. 

Then shall the cone FHK be equal to the rhombus. 


Construct a third cone ΗΝ with base (the circle about 
MN) equal to the base of OAB and height LP equal to OD. 


Then, since oP Om 
LP Cp OD On), 
But [Lemma 1] OD: CD=(rhombus OADB): (cone DAB), 
and IP :CD=(cone LMN): (cone DAB). 
It follows that 
(rhombus OA DB)= (cone LMI ).........s0ccc00s000 {1} 


ON THE SPHERE AND CYLINDER I. 25 


Again, since AB= MN, and 
(surface of OA B) =(base of FHK), 
(base of FHK) : (base of LMI) 
= (surface of OAB): (base of OA B) 
= 5 + BC [Prop. 15] 
=OD: DE, by similar triangles, 
= IP : FG, by hypothesis. 
Thus, in the cones FAK, LMN, the bases are reciprocally 
proportional to the heights. 
Therefore the cones FHK, LMN are equal, 


and hence, by (1), the cone FHK is equal to the given 
solid rhombus. 


Proposition 19. 


If an isosceles cone be cut by a plane parallel to the base, 
and on the resulting circular section a cone be described having 
as its apex the centre of the base [of the first cone], and if the 
rhombus so formed be taken away from the whole cone, the part 
remaining will be equal to the cone with base equal to the surface 
of the portion of the first cone between the parallel planes and 
with height equal to the perpendicular drawn from the centre of 
the base of the first cone on one side of that cone. 


Let the cone OAB be cut by a plane parallel to the base in 
the circle on DE as diameter. Let C be the centre of the base 
of the cone, and with C as apex and the circle about DE as base 


describe a cone, making with the cone ODE the rhombus 
ODCE. 


Take a cone FGH with base equal to the surface of the 
frustum DABE and height equal to the perpendicular (CK) 
from C on AO. 


Then shall the cone FGH be equal to the difference between 
the cone OAB and the rhombus ODCE. 


Take (1) a cone LMN with base equal to the surface of the 
cone OAB, and height equal to CK, 


26 ARCHIMEDES 


(2) acone PQR with base equal to the surface of the cone 
ODE and height equal to CK. 


ο 
ΖΞ 
Ξ) Ε G H 
K 
A Cc B 
L P 
N ἢ R 


M 


Now, since the surface of the cone OAB is equal to the 
surface of the cone ODE together with that of the frustum 
DABE, we have, by the construction, 

(base of LMN) = (base of FGH) + (base of PQR) 
and, since the heights of the three cones are equal, 
(cone LMN) =(cone FGH) + (cone PQR). 

But the cone ΖΗ͂Ν is equal to the cone OAB [Prop. 17], 
and the cone PQR is equal to the rhombus ODCE [Prop. 18]. 

Therefore (cone 0A B)=(cone FGH) + (rhombus ODCE), 


and the proposition is proved. 


Proposition 20. 


If one of the two isosceles cones forming a rhombus be cut 
by a plane parallel to the base and on the resulting circular 
section a cone be described having the same apex as the second 
cone, and if the resulting rhombus be taken from the whole 
rhombus, the remainder will be equal to the cone with base equal 
to the surface of the portion of the cone between the parallel 
planes and with height equal to the perpendicular drawn from 
the apex of the second * cone to the side of the first cone. 

* There is a slight error in Heiberg’s translation ‘‘ prioris coni” and in the 


corresponding note, p. 93. The perpendicular is not drawn from the apex of 
the cone which is cut by the plane but from the apex of the other. 


ON THE SPHERE AND CYLINDER I. ἢ 


Let the rhombus be OACB, and let the cone OAB be cut 
by a plane parallel to its base in the circle about D£ as diameter. 
With this circle as base and Cas apex describe a cone, which 
therefore with ODE forms the rhombus ODCE. 


AA 


Take a cone FGH with base equal to the surface of the 
frustum DABE and height equal to the perpendicular (Ck’) 
from C on OA. 

The cone FGH shall be equal to the difference between the 
rhombi OACB, ODCE. 

For take (1) a cone ZMN with base equal to the surface of 
OAB and height equal to CK, 

(2) a cone PQR, with base equal to the surface of ODE, 
and height equal to CK. 

Then, since the surface of OAB is equal to the surface of 
ODE together with that of the frustum DABE, we have, by 
construction, 

(base of LMN) = (base of PQR)+ (base of FGH), 
and the three cones are of equal height ; 
therefore (cone ZMN)=(cone PQR)+(cone FGH). 

But the cone LMN is equal to the rhombus OACB, and the 

cone PQR is equal to the rhombus ODCE [Prop. 18]. 


Hence the cone FGH is equal to the difference between the 
two rhombi OACB, ODCE. 


28 ARCHIMEDES 


Proposition 21. 


A regular polygon of an even number of sides being inscribed 
in a circle, as ABC...A'...C’B’A, so that AA’ is a diameter, 
uf two angular points next but one to each other, as B, B’, be 
joined, and the other lines parallel to BB' and joining pairs 
of angular points be drawn, as CC’, DD’..., then 

(BB'+CC'+...): AA’ =A’B: BA. 

Let BB’, CC’, DD’... meet AA’ in F, G, H....>amdmen 

CB’, DC’... be joined meeting AA’ in K, L,... respectively. 


Then clearly CB’, DC’,... are parallel to one another and to 
AB. 


Hence, by similar triangles, 
BFA — i pe 


= PRIAS 


ON THE SPHERE AND CYLINDER I. 29 


and, summing the antecedents and consequents respectively, we 
have 
(BB'+CC'+...): AA’= BF: FA 
= A’B: BA. 


Proposition 22. 


If a polygon be inscribed in a segment of a circle LAL’ so 
that all its sides eacluding the base are equal and their number 
even, as LK...A...K'L’, A being the middle point of the segment, 
and if the lines BB’, CC’,... parallel to the base LL’ and joining 
purrs of angular points be drawn, then 


(BB+ C0’ +...+ LM): AM=A'B: BA, 


where M is the middle point of LL’ and AA’ is the diameter 
through M. 


D 
o———__K 


ne UL 


Joiming CB’, DC’,...LK’, as in the last proposition, and 
supposing that they meet AM in P, Q,...R, while BB’, CC’...., 
KK’ meet AM in F, G,... H, we have, by similar triangles, 

BF: FA=B'F:FP 


—LM: RM; 


30 ARCHIMEDES 


and, summing the antecedents and consequents, we obtain 
(BB’+CC'+...4+£M):AM=BF: FA 
=A’B: BA. 


Proposition 23. 


Take a great circle ABC... of a sphere, and inscribe in it 
a regular polygon whose sides are a multiple of four in number. 
Let AA’, MM’ be diameters at right angles and joining 
opposite angular points of the polygon. 


Then, if the polygon and great circle revolve together about 
the diameter AA’, the angular points of the polygon, except A, 
A’, will describe circles on the surface of the sphere at right 
angles to the diameter AA’. Also the sides of the polygon 
will describe portions of conical surfaces, e.g. BC will describe 
a surface forming part of a cone whose base is a circle about 
CC’ as diameter and whose apex is the point in which CB, 
C’B’ produced meet each other and the diameter AA’. 


Comparing the hemisphere MAM’ and that half of the 
figure described by the revolution of the polygon which is 
included in the hemisphere, we see that the surface of the 
hemisphere and the surface of the inscribed figure have the 
same boundaries in one plane (viz. the circle on MM’ as 


ON THE SPHERE AND CYLINDER I. 91 


diameter), the former surface entirely includes the latter, and 
they are both concave in the same direction. 

Therefore [Asswmptions, 4] the surface of the hemisphere 
is greater than that of the inscribed figure; and the same is 
true of the other halves of the figures. 

Hence the surface of the sphere is greater than the surface 
described by the revolution of the polygon inscribed in the great 
circle about the diameter of the great circle. 


Proposition 24. 


If a regular polygon AB...A’...B’A, the number of whose 
sides is a multiple of four, be inscribed in a great circle of a 
sphere, and if BB’ subtending two sides be joined, and all the 
other lines parallel to BB’ and joimng pairs of angular points 
be drawn, then the surface of the figure inscribed in the sphere 
by the revolution of the polygon about the diameter AA’ is equal 
to a circle the square of whose radius is equal to the rectangle 

BA (BB’+ CC’ +...). 

The surface of the figure is made up of the surfaces of parts 

of different cones. 


Now the surface of the cone ABB’ is equal to a circle whose 
radius is VBA .4BB’, [Prop. 14] 


ἜΝ ARCHIMEDES 


The surface of the frustum BB’C’C is equal to a circle of 
radius VBC. 4(BB’ + CC’), [Prop. 16] 
and so on. 

It follows, since BA = BC=..., that the whole surface is 
equal to a circle whose radius is equal to 


NBA (BB τ UM's... £VY%, 


Proposition 25. 


The surface of the figure inscribed in a sphere as in the last 
propositions, consisting of portions of conical surfaces, is less than 
four times the greatest circle in the sphere. 

Let AB...A’...B’A be a regular polygon inscribed in a 


great circle, the number of its sides being a multiple of four. 


Mt 


As before, let BB’ be drawn subtending two sides, and 
CC’"...YY’ parallel to BB’. 
Let R be a circle such that the square of its radius is equal 


to 
AB(BB’+ C0’ +...+ VY’, 


so that the surface of the figure inscribed in the sphere is equal 
to Rf. [Prop. 24] 


ON THE SPHERE AND CYLINDER I. 33 


Now 
(BB + CC" + ...+ YY’): AA’=A’B: AB, [Prop. 21] 
whence AB(BB’+CC’+...4+ YY’)=AA’.A’B. 
Hence (radius of R)?’= AA’. A’B 
AA”: 


Therefore the surface of the inscribed figure, or the circle R, 
is less than four times the circle AMA’M’. 


Proposition 26. 


The figure inscribed as above in a sphere ts equal [in volume] 
to a cone whose base is a circle equal to the surface of the figure 
imscribed in the sphere and whose height is equal to the 
perpendicular drawn from the centre of the sphere to one side of 
the polygon. 


Suppose, as before, that AB...A’...B’A is the regular 
polygon inscribed in a great circle, and let BB’, CC’,... be 
joined. 


With apex O construct cones whose bases are the circles 
on BB’, CO’,... as diameters in planes perpendicular to AA’. 


H. A. 3 


34 ARCHIMEDES 


Then OBAB’ is a solid rhombus, and its volume is equal to 
a cone whose base is equal to the surface of the cone ABB’ and 
whose height is equal to the perpendicular from O on AB 
[Prop. 18]. Let the length of the perpendicular be p. 


Again, if CB, C’B’ produced meet in 7’, the portion of the 
solid figure which is described by the revolution of the triangle 
BOC about AA’ is equal to the difference between the rhombi 
OCTC' and OBTB’, 1.6. to a cone whose base is equal to the 
surface of the frustum BB’C’C and whose height is p [Prop. 20]. 


Proceeding in this manner, and adding, we prove that, since 
cones of equal height are to one another as their bases, the 
volume of the solid of revolution is equal to a cone with height 
p and base equal to the sum of the surfaces of the cone BAB’, 
the frustum BB’C’C, etc., i.e. a cone with height p and base 
equal to the surface of the solid. 


Proposition 27. 


The figure inscribed in the sphere as before is less than 
four times the cone whose base is equal to a great circle of 
the sphere and whose height is equal to the radius of the 
sphere. 

By Prop. 26 the volume of the solid figure is equal to a cone 
whose base is equal to the surface of the solid and whose height 
is p, the perpendicular from O on any side of the polygon. Let 
R be such a cone. 

Take also a cone S with base equal to the great circle, and 
height equal to the radius, of the sphere. 


Now, since the surface of the inscribed solid is less than four 
times the great circle [Prop. 25], the base of the cone £ is less 
than four times the base of the cone S. 

Also the height (p) of R is less than the height of S, 


Therefore the volume of {ὦ is less than four times that of S; 
and the proposition is proved. 


ON THE SPHERE AND CYLINDER I. 35 


Proposition 28. 


Let a regular polygon, whose sides are a multiple of four in 
number, be circumscribed about a great circle of a given 
sphere, as AB...A’...B’A; and about the polygon describe 
another circle, which will therefore have the same centre as the 
great circle of the sphere. Let AA’ bisect the polygon and 
cut the sphere in a, αἰ. 


If the great circle and the circumscribed polygon revolve 
together about AA’, the great circle will describe the surface 
of a sphere, the angular points of the polygon except A, A’ will 
move round the surface of a larger sphere, the points of contact 
of the sides of the polygon with the great circle of the inner 
sphere will describe circles on that sphere in planes perpen- 
dicular to AA’, and the sides of the polygon themselves will 
describe portions of conical surfaces. The circumscribed figure 
will thus be greater than the sphere itself. 

Let any side, as BM, touch the inner circle in K, and let K’ 
be the point of contact of the circle with B’M’. 

Then the circle described by the revolution of KK’ about 
AA’ is the boundary in one plane of two surfaces 


(1) the surface formed by the revolution of the circular 
segment Kak’, and 


3—2 


36 ARCHIMEDES 


(2) the surface formed by the revolution of the part 
KB...A...BK’ of the polygon. 

Now the second surface entirely includes the first, and they 
are both concave in the same direction ; 

therefore [Assumptions, 4] the second surface is greater 
than the first. 

The same is true of the portion of the surface on the opposite 
side of the circle on KK’ as diameter. 

Hence, adding, we see that the surface of the figure 
circumscribed to the given sphere is greater than that of the 
sphere itself. 


Proposition 29. 


In a figure circumscribed to a sphere in the manner shown 
in the previous proposition the surface 1s equal to a circle the 
square on whose radius is equal to AB(BB’ + CC’ +...). 

For the figure circumscribed to the sphere is inscribed in a 
larger sphere, and the proof of Prop. 24 applies. 


Proposition 30. 


The surface of a figure circumscribed as before about a sphere 
as greater than four times the great circle of the sphere. 


ON THE SPHERE AND CYLINDER I, 37 


Let AB...A’...B'A be the regular polygon of 4n sides 
which by its revolution about AA’ describes the figure circum- 
scribing the sphere of which ama'm’ is a great circle. Suppose 
aa’, AA’ to be in one straight line. 

Let R be a circle equal to the surface of the circumscribed 
solid. 

Now (BB’+CC’+...): AA'=A'B: BA, [asin Prop. 21] 
so that AB (BB'+C0"’+...)=AA’.A’B. 

Hence (radius of R)=V AA’. A’B [Prop. 29] 

> AB; 

But A’B = 20P, where P is the point in which AB touches 
the circle ama'm’. 

Therefore (radius of R) > (diameter of circle ama'm’) ; 


whence R, and therefore the surface of the circumscribed solid, 
is greater than four times the great circle of the given sphere. 


Proposition 91. 


The solid of revolution circumscribed as before about a sphere 
is equal to a cone whose base is equal to the surface of the sold 
and whose height is equal to the radius of the sphere. 

The solid is, as before, a solid inscribed in a larger sphere ; 
and, since the perpendicular on any side of the revolving polygon 
is equal to the radius of the inner sphere, the proposition is 
identical with Prop. 26. 


Cor. The solid circumscribed about the smaller sphere 1s 
greater than four times the cone whose base is a great circle 
of the sphere and whose height is equal to the radius of the 
sphere. 

For, since the surface of the solid is greater than four times 
the great circle of the inner sphere [Prop. 30], the cone whose 
base is equal to the surface of the solid and whose height is the 
radius of the sphere is greater than four times the cone of 
the same height which has the great circle for base. [Lemma 1.] 

Hence, by the proposition, the volume of the solid is greater 
than four times the latter cone. 


38 ARCHIMEDES 


Proposition 32. 


If a regular polygon with 4n sides be inscribed in a great 
circle of a sphere, as ab...a’...b’a, and a similar polygon 
AB...A’...B’A be described about the great circle, and if the 
polygons revolve with the great circle about the diameters aa’, 
AA’ respectively, so that they describe the surfaces of solid 
Jigures inscribed in and circumscribed to the sphere respectively, 
then 

(1) the surfaces of the circumscribed and inscribed figures 
are to one another in the duplicate ratio of their sides, and 

(2) the figures themselves [t.e. their volumes] are in the 
triplicate ratio of their sides. 


(1) Let AA’, aa’ be in the same straight line, and let 
MmOm'M’ be a diameter at right angles to them. 


Join BB’, CC’,... and bb’, cc’, ... which will all be parallel 
to one another and MM", 


Suppose R, S to be circles such that 
f =(surface of circumscribed solid), 


S =(surface of inscribed solid). 


ON THE SPHERE AND CYLINDER I. 39 


Then (radius of R)? = AB(BB’ + CC’ +...) [Prop. 29] 
(radius of 5} = αὖ (bb’ + cc’ + ...). [Prop. 24] 


And, since the polygons are similar, the rectangles in these 
two equations are similar, and are therefore in the ratio of 


AB” save, 
Hence 
(surface of circumscribed solid) : (surface of inscribed solid) 
=e 0 


(2) Take a cone V whose base is the circle R and whose 
height is equal to Oa, and a cone W whose base is the circle S 
and whose height is equal to the perpendicular from Ὁ on ab, 
which we will call p. 


Then V, W are respectively equal to the volumes of the 
circumscribed and inscribed figures. [Props. 31, 26] 


Now, since the polygons are similar, 
AB:ab=Oa:p 
= (height of cone V) : (height of cone W); 


and, as shown above, the bases of the cones (the circles R, 3) 
are in the ratio of AB? to ab’. 


Therefore V:W=AB?: ab*. 


Proposition 33S. 


The surface of any sphere is equal to four times the greatest 
circle in τί. 


Let C be a circle equal to four times the great circle. 


Then, if C is not equal to the surface of the sphere, it must 
either be less or greater. 


I. Suppose C less than the surface of the sphere. 


It is then possible to find two lines £, γ, of which β is the 
greater, such that 


B : y<(surface of sphere) : C. [Prop. 2] 


Take such lines, and let ὃ be a mean proportional between 
them. 


40 ARCHIMEDES 


Suppose similar regular polygons with 4n sides circum- 
scribed about and inscribed in a great circle such that the ratio 
of their sides is less than the ratio 8 : δ. [Prop. 3] 


Let the polygons with the circle revolve together about 
a diameter common to all, describing solids of revolution as 
before. 
Then (surface of outer solid) : (surface of inner solid) 
= (side of outer)’: (side of inner)’ [Prop. 32] 
<9? +105 ἡ 8 tiv 
< (surface of sphere) : C, a fortiori. 
But this is impossible, since the surface of the circum- 


scribed solid is greater than that of the sphere [Prop. 28], while 
the surface of the inscribed solid is less than (Οὗ [Prop. 25]. 


Therefore C is not less than the surface of the sphere. 


II. Suppose C greater than the surface of the sphere. 
Take lines β, γ, of which β is the greater, such that 
8:y<C: (surface of sphere). 


Circumscribe and inscribe to the great circle similar regular 
polygons, as before, such that their sides are in a ratio less than 
that of 8 to 8, and suppose solids of revolution generated in the 
usual manner. 


ON THE SPHERE AND CYLINDER I. 41 


Then, in this case, 

(surface of circumscribed solid) : (surface of inscribed solid) 
< C’: (surface of sphere). 

But this is impossible, because the surface of the circum- 
seribed solid is greater than C [Prop. 30], while the surface of 
the inscribed solid is less than that of the sphere [Prop. 23]. 

Thus C is not greater than the surface of the sphere. 

Therefore, since it is neither greater nor less, Cis equal to 
the surface of the sphere. 


Proposition 34. 


Any sphere is equal to four times the cone which has its base 
equal to the greatest circle in the sphere and its height equal 
to the radius of the sphere. 


Let the sphere be that of which ama’m’ is a great circle. 

If now the sphere is not equal to four times the cone 
described, it is either greater or less, 

I. If possible, let the sphere be greater than four times the 
cone. 


Suppose V to be a cone whose base is equal to four times 
the great circle and whose height is equal to the radius of the 
sphere. 


Then, by hypothesis, the sphere is greater than V; and two 
lines 8, y can be found (of which β is the greater) such that 
8B :y<(volume of sphere) : V. 


Between 8 and y place two arithmetic means ὃ, ε. 


As before, let similar regular polygons with sides 4n in 
number be circumscribed about and inscribed in the great 
circle, such that their sides are in a ratio less than B: 6. 


Imagine the diameter aa’ of the circle to be in the same 
straight line with a diameter of both polygons, and imagine 
the latter to revolve with the circle about aa’, describing the 


42 ARCHIMEDES 


surfaces of two solids of revolution. The volumes of these solids 
are therefore in the triplicate ratio of their sides. __[Prop. 32] 


Thus (vol. of outer solid) : (vol. of inscribed solid) 
< 6° : δ᾽, by hypothesis, 
<8:¥, a fortiori (since β : y> P* : δ", 
< (volume of sphere) : V, a fortiore. 


But this is impossible, since the volume of the circumscribed 


8 


* That B:y>*:6* is assumed by Archimedes, 


Eutocius proves the 
property in his commentary as follows. 


Take x such that (33 OS0 8 & 
Thus B-6:B=6-2:6 
and, since B>6, B-é6>6-2. 
But, by hypothesis, B-6=6-e. 
Therefore d6-e>6-a@, 
or τε 
Again, suppose One — Ξ ΠΣ 


and, as before, we have 

so that, a fortiori, 
Therefore 

and, since 23» ε, y>y. 


Now, by hypothesis, 8, δ, x, y are in continued proportion ; 
therefore Bo — peg 


Se} Ὑ- 


ON THE SPHERE AND CYLINDER I. 43 


solid is greater than that of the sphere [Prop. 28], while the 
volume of the inscribed solid is less than V [Prop. 27]. , 


Hence the sphere is not greater than V, or four times the 
cone described in the enunciation. 


II. If possible, let the sphere be less than Κ΄. 
In this case we take 8, y (8 being the greater) such that 
B:y< V: (volume of sphere). 


The rest of the construction and proof proceeding as before, 
we have finally 


(volume of outer solid) : (volume of inscribed solid) 
< V : (volume of sphere). 


But this is impossible, because the volume of the outer 
solid is greater than V [Prop. 31, Cor.], and the volume of the 
inscribed solid is less than the volume of the sphere. 


Hence the sphere is not less than JV. 


Since then the sphere is neither less nor greater than V, it 
is equal to V, or to four times the cone described in the enun- 
ciation. 


Cor. From what has been proved it follows that every 
cylinder whose base is the greatest circle in a sphere and whose 
height is equal to the diameter of the sphere is 3 of the sphere, 
and its surface together with its bases 1s 3 of the surface of the 
sphere. 


For the cylinder is three times the cone with the same 
base and height [Eucl. x1. 10], i.e. six times the cone with 
the same base and with height equal to the radius of the 
sphere. 


But the sphere is four times the latter cone [Prop. 34]. 
Therefore the cylinder is 3 of the sphere. 


Again, the surface of a cylinder (excluding the bases) is 
equal to a circle whose radius is a mean proportional between 
the height of the cylinder and the diameter of its base 
[Prop. 13]. 


44 ARCHIMEDES 


In this case the height is equal to the diameter of the base 
and therefore the circle is that whose radius is the diameter of 
the sphere, or a circle equal to four times the great circle of 
the sphere. 

Therefore the surface of the cylinder with the bases is equal 
to six times the great circle. 


And the surface of the sphere is four times the great circle 
[Prop. 33]; whence 


(surface of cylinder with bases) = 3. (surface of sphere). 


Proposition 35. 


If in a segment of a circle LAL’ (where A is the middle 
point of the arc) a polygon LK...A...K'L’ be inscribed of which 
LL’ is one side, while the other sides are 2n in number and all 
equal, and if the polygon revolve with the segment about the 
diameter AM, generating a solid figure inscribed in a segment of 
a sphere, then the surface of the inscribed solid is equal to a 
circle the square on whose radius is equal to the rectangle 

AB (BB'+C0' +... Pi 


A 


A’ 


The surface of the inscribed figure is made up of portions of 
surfaces of cones. 


ON THE SPHERE AND CYLINDER I. 45 


If we take these successively, the surface of the cone BAB 4 
is equal to a circle whose radius is 


VAB.4BB’. [Prop. 14] 
The surface of the frustum of a cone BCC’B’ is equal to 
a circle whose radius ᾿ 


BB’ — ae 


oe 


Proceeding in this way and adding, we find, since circles 
are to one another as the squares of their radii, that the 
surface of the inscribed figure is equal to a circle whose radius 
15 


[Prop. 16] 


and so on. 


ν 4B (BB +00" +... + KK' + 75°), 


Proposition 96. 


The surface of the figure inscribed as before in the segment 
of a sphere is less than that of the segment of the sphere. 

This is clear, because the circular base of the segment is a 
common boundary of each of two surfaces, of which one, the 
segment, includes the other, the solid, while both are concave 
in the same direction [Asswmptions, 4}. 


Proposition 37. 


The surface of the solid figure inscribed in the segment of the 
sphere by the revolution of LK...A...K’L’' about AM is less than 
a circle with radius equal to AL. 


Let the diameter AM meet the circle of which LAL’ is a 
segment againin A’, Join A’B. 

As in Prop. 35, the surface of the inscribed solid is equal to 
a circle the square on whose radius is 


AB(BR’ #00! τὸ K 4 DM). 


46 ARCHIMEDES 


But this rectangle =A BAM [Prop. 22] 
A αι AM 
< AL’. 


4 


A 


Hence the surface of the inscribed solid is less than the 
circle whose radius is AJL. 


Proposition 38. 


The solid figure described as before in a segment of a sphere 
less than a hemisphere, together with the cone whose base is the 
base of the segment and whose apex ts the centre of the sphere, 
ws equal to a cone whose base is equal to the surface of the 
inscribed solid and whose height is equal to the perpendicular 
Srom the centre of the sphere on any side of the polygon. 

Let O be the centre of the sphere, and p the length of the 
perpendicular from O on AB. 

Suppose cones described with O as apex, and with the 
circles on BB’, CC’,... as diameters as bases. 


Then the rhombus OBAB' is equal to a cone whose base is 
equal to the surface of the cone BAB’, and whose height is p. 

[Prop. 18] 

Again, if CB, C’B’ meet in T, the solid described by the 

triangle BOC as the polygon revolves about AO is the difference 


ON THE SPHERE AND CYLINDER I. 47 


between the rhombi OCTC’ and OBTB’, and is therefore equal 
to a cone whose base is equal to the surface of the frustum 
BCC’B' and whose height is p. [Prop. 20] 

Similarly for the part of the solid described by the triangle 
COD as the polygon revolves ; and so on. 


Ο 


Hence, by addition, the solid figure inscribed in the segment 
together with the cone OLL’ is equal to a cone whose base is 
the surface of the inscribed solid and whose height is p. 


Cor. The cone whose base 15 a circle with radius equal to 
AL and whose height is equal to the radius of the sphere is 
greater than the sum of the inscribed solid and the cone OLL’. 


For, by the proposition, the inscribed solid together with 
the cone OLL’ is equal to a cone with base equal to the surface 
of the solid and with height p. 


This latter cone is less than a cone with height equal to OA 
and with base equal to the circle whose radius is AL, because 
the height p is less than OA, while the surface of the solid is 
less than a circle with radius AJL. [ Prop. 37] 


Proposition 39. 


Let lal’ be a segment of a great circle of a sphere, being less 
than a semicircle. Let O be the centre of the sphere, and join 
Ol, Ol’. Suppose a polygon circumscribed about the sector Olal’ 
such that its sides, excluding the two radii, are 2n in number 


48 ARCHIMEDES 


and all equal, as LK,... BA, AB’,... K'L’; and let OA be that 
radius of the great circle which bisects the segment lal’. 

The circle circumscribing the polygon will then have the 
same centre O as the given great circle. 

Now suppose the polygon and the two circles to revolve 
together about OA. The two circles will describe spheres, the 


angular points except A will describe circles on the outer 
sphere, with diameters BB’ etc., the points of contact of the 
sides with the inner segment will describe circles on the inner 
sphere, the sides themselves will describe the surfaces of cones 
or frusta of cones, and the whole figure circumscribed to the 
segment of the inner sphere by the revolution of the equal 
sides of the polygon will have for its base the circle on LL’ 
as diameter. 

The surface of the solid figure so circumscribed about the 
sector of the sphere [excluding its base] will be greater than that 
of the segment of the sphere whose base is the circle on Il’ as 
diameter. 

For draw the tangents IT, 1’T’ to the inner segment at 1, 1’. 
These with the sides of the polygon will describe by their 
revolution a solid whose surface is greater than that of the 
segment [Asswmptions, 4]. 

But the surface described by the revolution of /7 is less 
than that described by the revolution of LT’, since the angle TUL 
is a right angle, and therefore LT >IT. 

Hence, a fortiori, the surface described by LK...A...K'L’ 
is greater than that of the segment. 


ON THE SPHERE AND CYLINDER I. 49 


Cor. The surface of the figure so described about the sector 
of the sphere is equal to a circle the square on whose radius 
is equal to the rectangle 

AB(BB’+C0'+...4+ KR'+4LL’), 

For the circumscribed figure is inscribed in the outer sphere, 

and the proof of Prop. 35 therefore applies. 


Proposition 40. 


The surface of the figure circumscribed to the sector as before 
is greater than a circle whose radius is equal to al. 

Let the diameter AaO meet the great circle and the circle 
circumscribing the revolving polygon again in α΄, A’. Join 
A’B, and let ON be drawn to JN, the point of contact of AB 
with the inner circle. 


Now, by Prop. 39, Cor., the surface of the solid figure 
circumscribed to the sector OLAI’ is equal to a circle the square 
on whose radius is equal to the rectangle 


AB (BR +00" KE 4). 


2 
But this rectangle is equal to A’B. AWM [as in Prop. 22]. 
Hi A. 4 


50 ARCHIMEDES 


Next, since AL’, al’ are parallel, the triangles AML’, aml’ 
are similar. And AL’><al’; therefore AM> am. 


Also A’B =20N =a’. 
Therefore A’B. AM>am.ad 
Sal”: 


Hence the surface of the solid figure circumscribed to the 
sector is greater than a circle whose radius is equal to al’, or al. 


Cor. 1. The volume of the figure circumscribed about the 
sector together with the cone whose apex is O and base the circle 
on LL’ as diameter, is equal to the volume of a cone whose base 
is equal to the surface of the circumscribed figure and whose 
height is ON. 

For the figure is inscribed in the outer sphere which has the 
same centre as the inner. Hence the proof of Prop. 38 applies. 


Cor. 2. The volume of the circumscribed figure with the cone 
OLL’ is greater than the cone whose base is a circle with radius 
equal to al and whose height is equal to the radius (Oa) of the 
inner sphere. 


For the volume of the figure with the cone OLL’ is equal to 
a cone whose base is equal to the surface of the figure and 
whose height is equal to ON. 


And the surface of the figure is greater than a circle with 
radius equal to al [Prop. 40], while the heights Oa, ON are 
equal. 


Proposition 41. 


Let lal’ be a segment of a great circle of a sphere which is 
less than a semicircle. 


Suppose a polygon inscribed in the sector Olal’ such that 
the sides lk,... ba, ab’,... kU are 2n in number and all equal. 
Let a similar polygon be circumscribed about the sector so that 
its sides are parallel to those of the first polygon; and draw 
the circle circumscribing the outer polygon. 

Now let the polygons and circles revolve together about 
OaA, the radius bisecting the segment lal’. 


ON THE SPHERE AND CYLINDER I. 51 


Then (1) the surfaces of the outer and inner solids of revolution 
so described are in the ratic of 4.8" to ab’, and (2) their volumes 
together with the corresponding cones with the same base and 
with apex O im each case are as 4.85 to ab’. 


(1) For the surfaces are equal to circles the squares on 
whose radii are equal respectively to 


AB (BB’+C0'+.... Le 
[Prop. 39, Cor.] 
and ab (υ +cec+...+kk' + sa) [Prop. 35] 


But these rectangles are in the ratio of AB? to ab’. Therefore 
so are the surfaces. 

(2) Let OnN be drawn perpendicular to ab and AB; and 
suppose the circles which are equal to the surfaces of the outer 
and inner solids of revolution to be denoted by S, s respectively. 

Now the volume of the circumscribed solid together with 
the cone OLL’ is equal to a cone whose base is S and whose 
height is ON [Prop. 40, Cor. 1]. 

And the volume of the inscribed figure with the cone Oll’ is 
equal to a cone with base s and height On [Prop. 38]. 

But S:s= AB’: ab’, 
and ON On —sA Bab, 

Therefore the volume of the circumscribed solid together with 
the cone OLTL’ is to the volume of the inscribed solid together 
with the cone Oll’ as AB? is to ab’ [Lemma 5]. 


4—2 


52 ARCHIMEDES 


Proposition 42. 


Tf lal' be a segment of a sphere less than a hemisphere and 
Oa the radius perpendicular to the base of the segment, the 
surface of the segment is equal to a circle whose radius is equal 
to al. 

Let R be a circle whose radius is equal to al. Then the 
surface of the segment, which we will call S, must, if it be not 
equal to R, be either greater or less than R. 


I. Suppose, if possible, S > 1. 


Let lal’ be a segment of a great circle which is less than a 
semicircle. Join Ol, Ol’, and let similar polygons with 2n equal 
sides be circumscribed and inscribed to the sector, as in the 
previous propositions, but such that 

(circumscribed polygon) : (inscribed polygon) < S: R. 
[Prop. 6] 
Let the polygons now revolve with the segment about OaJA, 


generating solids of revolution circumscribed and inscribed to 
the segment of the sphere. 


Then 
(surface of outer solid) : (surface of inner solid) 
=A ἢ [Prop. 41] 
= (circumscribed polygon) : (inscribed polygon) 
<S: R, by hypothesis. 
But the surface of the outer solid is greater than S [Prop. 39]. 


ON THE SPHERE AND CYLINDER 1. 53 


Therefore the surface of the inner solid is greater than R; 
which is impossible, by Prop. 37. 

II. Suppose, if possible, S< R. 

In this case we circumscribe and inscribe polygons such that 
their ratio is less than R: S; and we arrive at the result that 

(surface of outer solid) : (surface of inner solid) 
<i: 8. 

But the surface of the outer solid is greater than & [Prop. 40]. 
Therefore the surface of the inner solid is greater than S: which 
is impossible [Prop. 36]. 

Hence, since S is neither greater nor less than R, 


Sie 


Proposition 43. 


Even of the segment of the sphere is greater than a henusphere, 
its surface is still equal to a circle whose radius is equal to al. 

For let lal’a’ be a great circle of the sphere, aa’ being the 
diameter perpendicular to J’; and let 


lav be a segment less than a semi- a! 
circle. 
Then, by Prop. 42, the surface of 1 7, 


the segment la’l’ of the sphere is 
equal to a circle with radius equal to 
αἹ. 

Also the surface of the whole 
sphere is equal to a circle with radius 
equal to aa’ [Prop. 33]. 

But aa’’?—a/l’?=al’, and circles are to one another as the 
squares on their radu. 

Therefore the surface of the segment lal’, being the difference 
between the surfaces of the sphere and of la’l’, is equal to a 
circle with radius equal to al. 


54 ARCHIMEDES 


Proposition 44. 


The volume of any sector of a sphere is equal to a cone whose 
base is equal to the surface of the segment of the sphere included 
in the sector, and whose height is equal to the radius of the 
sphere. 

Let R be a cone whose base is equal to the surface of the 
segment lal’ of a sphere and whose height is equal to the radius 
of the sphere; and let S be the volume of the sector Olal’. 


β 
ὃ 
ε 
oY 


Then, if S is not equal to R, it must be either greater or 
less. 


I. Suppose, if possible, that S> R. 


Find two straight lines B, y, of which @ is the greater, such 
that 
Bry τ. 
and let 6, ε be two arithmetic means between 8, γ. 


Let lal’ be a segment of a great circle of the sphere. 
Join Ol, Ol’, and let similar polygons with 2n equal sides be 
circumscribed and inscribed to the sector of the circle as before, 
but such that their sides are in a ratio less than #: ὃ. 
[Prop. 4]. 


ON THE SPHERE AND CYLINDER I. δῦ 


Then let the two polygons revolve with the segment about 
OaA, generating two solids of revolution. 


Denoting the volumes of these solids by V, v respectively, 
we have 


(V +cone OLL’) : (v+cone Oll’)= 4.5": αν" [Prop. 41] 
</B> Ὁ δὲ 
<8:y¥, ὦ fortior™, 
«Κ᾽: R, by hypothesis. 
Now (V+cone OLL’)>S. 
Therefore also (v+cone Ol’) > R. 
But this is impossible, by Prop. 38, Cor. combined with Props. 
42, 43. 
Hence S>R. 
II. Suppose, if possible, that S< R. 
In this case we take 8, y such that 
Biya s, 
and the rest of the construction proceeds as before. 
We thus obtain the relation 
(V+ cone OLL’'):(v+ cone Oll’)< R:S. 
Now (v+cone Ol’) <8. 
Therefore (V+cone OLL’)< R; 
which is impossible, by Prop. 40, Cor. 2 combined with Props. 
42, 48. 
Since then S is neither greater nor less than R, 


San 


* Cf. note on Prop. 34, p. 42.. 


ON THE SPHERE AND CYLINDER. 


BOOK II. 


“ ARCHIMEDES to Dositheus greeting. 


On a former occasion you asked me to write out the proofs of 
the problems the enunciations of which I had myself sent to 
Conon. In point of fact they depend for the most part on the 
theorems of which I have already sent you the demonstrations, 
namely (1) that the surface of any sphere is four times the 
greatest circle in the sphere, (2) that the surface of any 
segment of a sphere is equal to a circle whose radius is equal 
to the straight line drawn from the vertex of the segment to 
the circumference of its base, (3) that the cylinder whose base 
is the greatest circle in any sphere and whose height is equal 
to the diameter of the sphere is itself in magnitude half as 
large again as the sphere, while its surface [including the two 
bases] is half as large again as the surface of the sphere, and 
(4) that any solid sector is equal to a cone whose base is the 
circle which is equal to the surface of the segment of the sphere 
included in the sector, and whose height is equal to the radius 
of the sphere. Such then of the theorems and problems as 
depend on these theorems I have written out in the book 
which I send herewith; those which are discovered by means 
of a different sort of investigation, those namely which relate 
to spirals and the conoids, I will endeavour to send you soon. 


ON THE SPHERE AND CYLINDER II. at 


The first of the problems was as follows: Given a sphere, to 
find a plane area equal to the surface of the sphere. 

The solution of this is obvious from the theorems aforesaid. 
For four times the greatest circle in the sphere is both a plane 
area and equal to the surface of the sphere. 


The second problem was the following.” 


Proposition 1. (Problem.) 


Given a cone or a cylinder, to find a sphere equal to the cone 
or to the cylinder. 


If V be the given cone or cylinder, we can make a cylinder 
equal to 3V. Let this cylinder be the cylinder whose base 
is the circle on AB as diameter and whose height is OD. 


Now, if we could make another cylinder, equal to the 
cylinder (OD) but such that its height is equal to the diameter 
of its base, the problem would be solved, because this latter 
cylinder would be equal to 3V, and the sphere whose diameter 
is equal to the height (or to the diameter of the base) of the 
same cylinder would then be the sphere required [I. 34, Cor.]. 


M 


Suppose the problem solved, and let the cylinder (CG) be 
equal to the cylinder (OD), while ZF, the diameter of the base, 
is equal to the height CG. 


58 ARCHIMEDES 


Then, since in equal cylinders the heights and bases are 
reciprocally proportional, 


ABE MEH CG: OD 


= + OD. vcs ce nese (1) 
Suppose MN to be such a line that 
TEEPE ΝΣ (2) 
Hence AB: HF=HEHEF: MN, 


and, combining (1) and (2), we have 
AB:MN= EF: OD, 
or ABH = MN = OD. 
Therefore AB: EF=EF:MN=MN : OD, 
and EF, MN are two mean proportionals between AB, OD. 


The synthesis of the problem is therefore as follows. Take 
two mean proportionals HF, MN between AB and OD, and 
describe a cylinder whose base is a circle on HF as diameter 
and whose height CG is equal to HF. 


Then, since 
AB: EF=EF:MN=MN: OD, 
EE? = AB MN, 
and therefore AB 2 hi — Ad May 
=i OD 
=CG:OD- 


whence the bases of the two cylinders (OD), (CG) are recipro- 
cally proportional to their heights. 


Therefore the cylinders are equal, and it follows that 
cylinder (CG) = 3V. 


The sphere on HF as diameter is therefore the sphere 
required, being equal to V. 


ON THE SPHERE AND CYLINDER II. 59 


Proposition 2. 


If BAB’ be a segment of a sphere, BB’ a diameter of the 
base of the segment, and O the centre of the sphere, and if AA’ 
be the diameter of the sphere bisecting BB’ in M, then the volume 
of the segment is equal to that of a cone whose base is the same 
as that of the segment and whose height 1s h, where 

h: AM=O0A’'+A'M: A'M. 

Measure MH along MA equal to h, and MH’ along MA’ 

equal to h’, where 
h’: A'M=0A+AM: AM. 

Suppose the three cones constructed which have 0, H 
H’ for their apices and the base (BB’) of the segment for their 
common base. Join AB, A’B. 


Let C be a cone whose base is equal to the surface of the 
segment BAB’ of the sphere, 1.6. to a circle with radius equal 
to AB [I. 42], and whose height is equal to OA. 

Then the cone (15 equal to the solid sector OBA B’ ΠΤ 44... 

Now, since ΠΗ :-ΜΑ - ΟΑΔ τ ΑΜ : ΑἸ, 


dividendo, HA: AM = OA Aw. 
and, alternately, HA: AO=AM: MA’, 
so that 
HO: OA=AA’:A'M 
= AB’: BM’ 


= (base of cone () : (circle on BB’ as diameter). 


60 ARCHIMEDES 


But OA is equal to the height of the cone C; therefore, since 
cones are equal if their bases and heights are reciprocally 
proportional, it follows that the cone C (or the solid sector 
OBAB’) is equal to a cone whose base is the circle on BB’ as 
diameter and whose height is equal to OH. 


And this latter cone is equal to the sum of two others 
having the same base and with heights OM, MH, 1.6. to the 
solid rhombus OBHB’. 


Hence the sector OBAB’ is equal to the rhombus OBHB’. 
Taking away the common part, the cone OBB’, 

the segment BAB’ = the cone HBB’. 
Similarly, by the same method, we can prove that 

the segment BA’B’ = the cone H’BB’. 
Alternative proof of the latter property. 


Suppose D to be a cone whose base is equal to the surface 
of the whole sphere and whose height is equal to OA. 


Thus D is equal to the volume of the sphere. [1]. 33, 34] 
Now, since OA’+ A’M: A’M=HM: MA, 
dividendo and alternando, as before, 
OA AH =A MMA: 
Again,since H’M: MA'’=O0A+AWM: AM, 
AC OA = AG ΠΗ 
= OA : AH, from above. 
Componendo, HO; OA OTe τ ΕΣ (1). 
Alternately, TO 2 Ol = 0 Ar eA Te iceecc sac neteeeeee (2), 
and, componendo, HH’: HO=OH: HA, 
=O OA wironn(d)): 


whence ΠῊ OA = ΗΠ τ (3). 
Next, since H’O: Ὁ —=OA% Athy (2); 
= AM: MA, 


(H'0+ 0H): H’0.0H=(A’M + MA)’ : A'M. MA, 


ON THE SPHERE AND CYLINDER II. 61 


whence, by means of (3), 
HH? Sp Om AA AM OMA, 
or rte OA — AA: BI: 
Now the cone D, which is equal to the sphere, has for its base 


a circle whose radius is equal to AA’, and for its height a line 
equal to OA. 


Hence this cone D is equal to a cone whose base is the circle 
on BB’ as diameter and whose height is equal to HH’; 


therefore the cone D =the rhombus HBH’B’, 

or the rhombus HBH’B’ = the sphere. 

But the segment BAB'= the cone HBB’; 
therefore the remaining segment BA’B’=the cone H’BB’. 


Cor. The segment BAB’ is to a cone with the same base and 
equal height in the ratio of OA'+ Α΄Η to A’M. 


Proposition 8. (Problem.) 


To cut a gwen sphere by a plane so that the surfaces of the 
segments may have to one another a given ratio. 


Suppose the problem solved. Let AA’ be a diameter of a 
great circle of the sphere, and suppose that a plane perpendicular 
to AA’ cuts the plane of the great circle in the straight 


B 


Β΄ 


line BB’, and AA’ in M, and that it divides the sphere so that 
the surface of the segment BAB’ has to the surface of the 
segment BA’B’ the given ratio. 


62 ARCHIMEDES 
Now these surfaces are respectively equal to circles with 
radu equal to AB, A’B [I. 42, 43]. 
Hence the ratio 4": A’B* is equal to the given ratio, 1.6, 
AWM is to MA’ in the given ratio. 
Accordingly the synthesis proceeds as follows. 
If H: Καὶ be the given ratio, divide AA’ in M so that 
AMMA =H 3K. 
Then AM : MA An Ae 
=(circle with radius AB) : (circle with radius A'B) 
= (surface of segment BAB’) : (surface of segment BA’B’). 


Thus the ratio of the surfaces of the segments is equal to 
the ratio ἢ : K. 


Proposition 4. (Problem.) 


To cut a given sphere by a plane so that the volumes of the 
segments are to one another in a given ratio. 

Suppose the problem solved, and let the required plane cut 
the great circle ABA’ at right angles in the line BB’. Let 
AA’ be that diameter of the great circle which bisects BB’ at 
right angles (in 1,7), and let O be the centre of the sphere. 


Take H on OA produced, and H’ on OA’ produced, such 
that 


OA’ A’ M = AM = Vege 2 eee aby 
and OA + AM: AM = HMMA oj ccccsets: (2). 
Jom BH. hod, BH BA”. 


ON THE SPHERE AND CYLINDER II. 63 


Then the cones HBB’, H’BB’ are respectively equal to the 
segments BAB’, BA’B’ of the sphere [Prop. 2]. 

Hence the ratio of the cones, and therefore of their altitudes, 
is given, Le. 

HM : H’M =the given ratio............... (3). 

We have now three equations (1), (2), (3), im which there 
appear three as yet undetermined points WM, H, H’; and it is 
first necessary to find, by means of them, another equation in 
which only one of these points (J/) appears, 1.86. we have, so to 
speak, to eliminate H, H’. 

Now, from (8), it is clear that HH’: H’M is also a given 
ratio; and Archimedes’ method of elimination is, first, to find 
values for each of the ratios A’H’: H’M and HH’: H’A’ which 
are alike independent of H, H’, and then, secondly, to equate 
the ratio compounded of these two ratios to the known value 


of the ratio HH’: H'M. 
(a) ‘To find such a value for A’H’: H’M. 
It is at once clear from equation (2) above that 
A MEO AOA AM esc tes. (4). 
(Ὁ) Τὸ find such a value for HH’: A’H’. 


From (1) we derive 


A’M: MA=04A'+A'M: HM 


HUAN PAUL ES ee sue ss ΝΟΣ δι τ (ΟἿ: 
and, from (2), A’ M:MA=H'M:0A+AM 
ta a ONO) |e ee τ ῊΣ (6). 
Thus HA: AO = OA ἼΣΗ 
whence OH 2? 04 =O Ava 
or OH OHS 0 Ab eAC re 


It follows that 
Wi OH — OE ΕΟ 
or EDEL OH Al One: 
Wherefore, HH’ ΗΠ Ἢ =O? eA 
= AA”: A’M’, by means of (6) 


64 ARCHIMEDES 


(c) Τὸ express the ratios A’H’: H’M and HH': H’M more 
simply we make the following construction. Produce OA to D 
so that OA=AD. (D will lie beyond H, for A’M> MA, and 
therefore, by (5), OA > AH.) 


Then ATE ENS OA OA + AM 
ΞΞ ADs) DM - τ τους τ ΕΣ (7). 
Now divide AD at E so that 
TPC dM =A DDE... caste (8). 


Thus, using equations (8), (7) and the value of HH’: H’A’ 
above found, we have 
ADs DEE Ed os’ ME 
9 (JEL ΕΟ 6 Bags hag NG ie oR £21) 
Meme a ἢ AE) (CAL ee 
But AD DE — (DM: DE)..(A Di. DM): 

Therefore MD IDE =A A” > AVM. ΝΣ (9). 
And D is given, since AD=OA. Also AD: DE (being equal 
to HH’: H'M) isa given ratio. Therefore D£ is given. 

Hence the problem reduces itself to the problem of diwiding 
A'D into two parts at M so that 

MD : (a given length) = (a given area) : 4΄}5. 

Archimedes adds: “If the problem is propounded in this 
general form, it requires a διορισμός [1.e. 1t 1s necessary to 
investigate the limits of possibility], but, if there be added the 
conditions subsisting in the present case, it does not require a 
διορισμός." 

In the present case the problem is: 

Given a straight line A’A produced to D so that A’'A =2AD, 
and given a point E on AD, to cut AA’ in a point M so that 

ACA MD DE. 

“And the analysis and synthesis of both problems will be 
given at the end*.” 

The synthesis of the main problem will be as follows. Let 
R:S be the given ratio, R being less than S. AA’ being a 


* See the note following this proposition. 


ON THE SPHERE AND CYLINDER II. 65 


diameter of a great circle, and O the centre, produce OA to D 
so that OA = AD, and divide AD in £ so that 


AB ED= fh eS. 
Then cut AA’ in UM so that 
VOD i — AA: AM, 
Through WM erect a plane perpendicular to 4A’; this plane 


will then divide the sphere into segments which will be to one 
another as ft to S. 


Take H on 4΄ 4 produced, and H’ on AA’ produced, so that 
Oil eA CAM FM MA a ince τ δὶ (1), 
OA eet ANE HM OMA’ πὸ Ὁ (2). 
We have then to show that 
HM: MH’=8:S, or AE: ED. 
(a) We first find the value of HH’: H’A’ as follows. 
As was shown in the analysis (0), 
AE SEALS OH *, 
or AE ae Oe SEP A” 
ΞΞ- 3 -A 
= MD: DE, by construction. 
(8) Next we have 
H’A’: H'M=0A:0A+ AM 
= AD: DM. 

Therefore HH’: H’M=(H@’: H’A’).(H'A’: H’M) 
=(MD: DE).(AD: DM) 
=A) ee 

whence HM: MH’=AE: ED 
— i, : ϑ. Gh, D: 


Note. The solution of the subsidiary problem to which the 
original problem of Prop. 4 is reduced, and of which Archimedes 
promises a discussion, is given in a highly interesting and 
important note by Eutocius, who introduces the subject with 
the following explanation. 


H. A. 5 


66 ARCHIMEDES 


“He [Archimedes] promised to give a solution of this 
problem at the end, but we do not find the promise kept in any 
of the copies. Hence we find that Dionysodorus too failed to 
light upon the promised discussion and, being unable to grapple 
with the omitted lemma, approached the original problem in a 
different way, which I shall describe later. Diocles also ex- 
pressed in his work περὶ πυρίων the opinion that Archimedes 
made the promise but did not perform it, and tried to supply 
the omission himself. His attempt I shall also give in its 
order. It will however be seen to have no relation to the 
omitted discussion but to give, like Dionysodorus, a construction 
arrived at by a different method of proof. On the other hand, 
as the result of unremitting and extensive research, I found in 
a certain old book some theorems discussed which, although the 
reverse of clear owing to errors and in many ways faulty as 
regards the figures, nevertheless gave the substance of what I 
sought, and moreover to some extent kept to the Doric dialect 
affected by Archimedes, while they retained the names familiar in 
old usage, the parabola being called a section of a right-angled 
cone, and the hyperbola a section of an obtuse-angled cone; 
whence I was led to consider whether these theorems might 
not in fact be what he promised he would give at the end. For 
this reason I paid them the closer attention, and, after finding 
great difficulty with the actual text owing to the multitude of 
the mistakes above referred to, I made out the sense gradually 
and now proceed to set it out, as well as I can, in more familiar 
and clearer language. And first the theorem will be treated 
generally, in order that what Archimedes says about the limits 
of possibility may be made clear; after which there will follow 
the special application to the conditions stated in his analysis 
of the problem.” 


The investigation which follows may be thus reproduced. 
The general problem is: 


Given two straight lines AB, AC and an area D, to divide 
AB at M so that 


AM:AC=D : MB’. 


ON THE SPHERE AND CYLINDER II. 67 


Analysis. 


Suppose JM found, and suppose AC placed at right angles to 
AB. Join CM and produce it. Draw HBN through B parallel 
to AC meeting CM in N, and through C draw CHE parallel to 
AB meeting HBN in E. Complete the parallelogram CEN F, 
and through M draw PMH parallel to AC meeting FN in P. 

Measure HL along EN so that 

GH: ΤΣ AB, HEY =D). ἢ - τ 3 
] 

Then, by hypothesis, 

AM: AC= CE. EE? MB". 
And 


AM: AC=CE: EN, A B 
by similar triangles, 
=CH. EL: EL. EN. 

It follows that PN’?=MB°=EL.EN. 


Hence, if a parabola be described with vertex ZH, axis HN, and 
parameter equal to HZ, it will pass through P; and it will be 
given in position, since HZ is given. 


Cc H [= 


Therefore P hes on a given parabola. 
Next, since the rectangles FH, AE are equal, 
FP PH=AB. BE. 


Hence, if a rectangular hyperbola be described with CH, CF 
as asymptotes and passing through B, it will pass through P. 
And the hyperbola is given in position. 


Therefore P lies on a given hyperbola. 


Thus P is determined as the intersection of the parabola 
and hyperbola. And since P is thus given, M is also given. 


᾿ διορισμός. 
Now, since AM:AC=D:MB* 
AM; MB*= ACD). 


But AC. Dis given, and tt will be proved later that the maximum 
value of AM. MB? is that which it assumes when BM = 2AM. 


5—2 


68 ARCHIMEDES 


Hence τ ts a necessary condition of the possibility of a 
solution that AC.D must not be greater than 1AB.(2AB)’, or 
fj AB 
27 7 

Synthesis. 


If O be such a point on AB that BO =2A0, we have seen 
that, in order that the solution may be possible, 


AG DAO OB". 
Thus AC. D is either equal to, or less than, AO. OB’. 


(1) IfAC.D=A0O. OB’, then the point Ο itself solves the 
problem. 


(2) Let AC.D be less than AO. OB”. 


Place AC at right angles to AB. Join CO, and produce it 
to R. Draw HBR through B parallel to AC meeting CO in R, 
and through C draw CE parallel 
to AB meeting HBR in Δ. Com- 
plete the parallelogram CHRF, 
and through O draw QOK parallel 
to AC meeting FR in Q and CH 
in K. 
Then, since 
ACD ΞΡ. 
measure RQ’ along RQ so that 
AC. D'=A0O O'R 
or AO eA 1a aes 
Measure Δ᾽}, along HR so that 
D=CE.EL (or AB. EL). 
Now, since AO: AC=D: Q’R’, by hypothesis, 
=CL BL Oar. 
and AO: AC=CE : ER, by similar triangles, 
= CH. EL: HL ER, 


it follows that 
OR? = EL. ER, 


ON THE SPHERE AND CYLINDER II. 69 


Describe a parabola with vertex H, axis HR, and parameter 
equal to HL. This parabola will then pass through Q’. 


Again, rect. FK = rect. AE, 
or FO. Qk — AS BE: 
and, if we describe a rectangular hyperbola with asymptotes 
CE, CF and passing through B, it will also pass through Q. 


Let the parabola and hyperbola intersect at P, and through 
P draw PMH parallel to AC meeting AB in M and CE 
in H, and GPN parallel to AB meeting CF in G and ER 
in NV. 


Then shall M be the required point of division. 
Since POPE = ABBE, 
rect. GM = rect. ME, 
and therefore CIN is a straight line. 


Thus AB? Pier G ΗΠ AMEN yee: (1). 
Again, by the property of the parabola, 
PN*=EL.EN, 

or I eee BLN ας τν τ ον stan (2). 


From (1) and (2) 
AM EL — AB. BE MB", 


or AM ABTAB EL —=AB OAC? MB. 
Alternately, 

AM AB: AGL AC=AB EL +: MB". 
or ANAC — D> MB* 


Proof of διορισμός. 


It remains to be proved that, if AB be divided at O so that 
BO =2A0, then AO. OB? is the maximum value of AM. MB’, 


or AO; OBS >AM MB. 
where M is any point on AB other than O. 


70 ARCHIMEDES 


Suppose that AO: AC=CEH.EL’: OB’, 
so that AO.0B*=CE.EL’. AC. 
Join CO, and produce it to V; 
draw HBN through B parallel “ 
to AC, and complete the paral- 
lelogram CE NF. 

Through O draw POH 
parallel to AC meeting FN 
iny-P and ΟἿ im ΤΗΣ 

With vertex H, axis EN, 
and parameter HL’, describe 
a parabola. This will pass 
through P, as shown in the 
analysis above, and beyond P α 
will meet the diameter CF of 
the parabola in some point. 

Next draw a rectangular 
hyperbola with asymptotes CZ, 
CF and passing through B. 
This hyperbola will also pass 
through P, as shown in the 
analysis. 

Produce VE to T so that 
TE=EN. Jom TP meeting 
CE in Y, and produce it to 
meet Cf in W. Thus 7'P will 
touch the parabola at P. 


Then, since BO=2A0; 
TP = 2P iW: 

And TP = QV 

Therefore Pie 


Since, then, WY between the asymptotes is bisected at P, the 
point where it meets the hyperbola, 


WY is a tangent to the hyperbola. 


Hence the hyperbola and parabola, having a common tangent 
at P, touch one another at P. 


ON THE SPHERE AND CYLINDER II. (al 


Now take any point M on AB, and through M draw ΟΥ̓Κ 
parallel to AC meeting the hyperbola in Q and CE in K. 
Lastly, draw GqQR through Q parallel to AB meeting CF in G, 
the parabola in q, and ΜΝ in R. 

Then, since, by the property of the hyperbola, the rectangles 
GK, AE are equal, CR is a straight line. 


By the property of the parabola, 


qk? = EL’. ER, 
so that QR? < EL’. ER. 
Suppose QR*=EL.ER, 


and we have AM:AC=CEH: ER 
=CH.EL:EL.ER 


=CH.EL: QR 
=CH.EL +: MB’, 
or AM.MB*=CE.EL.AC. 
Therefore AM.MB’<CE.EL’'.AC 
<< AAO ON ose 


If AC. D< AO. OB’, there are two solutions because there 
will be two points of intersection between the parabola and the 
hyperbola. 

For, if we draw with vertex # and axis HN a parabola 
whose parameter is equal to #Z, the parabola will pass through 
the point ᾧ (see the last figure); and, since the parabola meets 
the diameter CF beyond Q, it must meet the hyperbola again 
(which has CF for its asymptote). 

(ii we put ΡΞ ἘΞ AC —c and )=b the pro- 
portion 

AM AC — DMB" 
is seen to be equivalent to the equation 
xv (a—2)=b'e, 
being a cubic equation with the term containing # omitted. 


Now suppose ΚΝ, EC to be axes of coordinates, HN being 


the axis of γ. 


72 ARCHIMEDES 


Then the parabola used in the above solution is the 
parabola 


and the rectangular hyperbola is 
y(a—2)= αὐ. 
Thus the solution of the cubic equation and the conditions 


under which there are no positive solutions, or one, or two 
positive solutions are obtained by the use of the two conics. ] 


[For the sake of completeness, and for their intrinsic interest, 
the solutions of the original problem in Prop. 4 given by 
Dionysodorus and Diocles are here appended. 


Dionysodorus’ solution. 


Let AA’ be a diameter of the given sphere. It is required 
to find a plane cutting AA’ at right angles (in a point M, 
suppose) so that the segments into which the sphere is divided 
are in a given ratio, as CD : DE. 


Produce A’A to F' so that AF’'= OA, where O is the centre 
of the sphere. 


Draw AH perpendicular to AA’ and of such length that 
FA :AH=CE: ED, 


ON THE SPHERE AND CYLINDER II. 73 


and produce AH to K so that 


With vertex F, axis FA, and parameter equal to AH 


describe a parabola. This will pass through K, by the equa- 
tion (a). 


Draw A’K’ parallel to AK and meeting the parabola in Kk’; 
and with A’F, A’K’ as asymptotes describe a rectangular 
hyperbola passing through H. This hyperbola will meet the 
parabola at some point, as P, between K and K’. 


Draw PM perpendicular to AA’ meeting the great circle in 
B, B’, and from H, P draw HL, PR both parallel to AA’ and 
meeting 4“ Κ΄ in L, αὶ respectively. 
Then, by the property of the hyperbola, 
PReE AH. HL, 


1. PM.MA'=HA.AA’, 
or PM:AH=AA’: A’'M, 
and ὌΠ. ΠΗ AAS = ACM". 
Also, by the property of the parabola, 
PM’ = FM. AH, 
iL€, FM; PM— PM: Ad, 
or PM Abe ees ΠῚ 


= AA”: A’M”’, from above. 


Thus, since circles are to one another as the squares of their 
radii, the cone whose base is the circle with A’ as radius and 
whose height is equal to #M, and the cone whose base is the 
circle with AA’ as radius and whose height is equal to AH, 
have their bases and heights reciprocally proportional. 


Hence the cones are equal; 1e., if we denote the first cone 
by the symbol ὁ (A’I/), FM, and so on, 


c(A'M), FM =c(AA’), AX. 
INow o(4 A), ΧΆ : CAG ea — EA): ALE 
=CE : ED, by construction. 


74 ARCHIMEDES 


Therefore 
¢(AA’), FA se Aw), YM = CE: ED ieee (8). 
But (1) c(AA’), FA =the sphere. [I. 34] 


(2) c(A’M), FM can be proved equal to the segment of 
the sphere whose vertex is A’ and height A’M. 


For take G on AA’ produced such that 
GM:MA'=FM: MA 


=0A+AM: AM. 
Then the cone GBB’ is equal to the segment A’BB’ [Prop. 2]. 
And FM: MG=AM: M4’, by hypothesis, 
= BM’: A’M* 
Therefore 
(circle with rad. BM) : (circle with rad. A’M) 
= FM: MG, 
so that c(A’M), FM =c (BM), MG 


= the segment A’BB’. 
We have therefore, from the equation (8) above, 
(the sphere) : (segmt. A’BB’) = CE : ED, 

whence (segmt. ABB’): (segmt. A’BB')=CD: DE. 


Diocles’ solution. 


Diocles starts, like Archimedes, from the property, proved in 
Prop. 2, that, if the plane of section cut a diameter AA’ of the 
sphere at right angles in 277, and if H, H’ be taken on OA, OA’ 
produced respectively so that 

OA’ 1 AM AM 1.5. λοι, 
OA+AM:AM=H’'M: MA’, 


then the cones HBB’, H’BB’ are respectively equal to the 
segments ABB’, A’BB’. 


ON THE SPHERE AND CYLINDER II. to 


Then, drawing the inference that 
HA :AM=0O4A’': A’'M, 
AA MOA, + AM, 


H von Oo A! 


Β' 


he proceeds to state the problem in the following form, slightly 
generalising it by the substitution of any given straight line for 
OAl-or OA": 

Given a straight line AA’, its extremities A, A’,a ratio C: D, 
and another straight line as AK, to divide AA’ at M and to find 
two points H, H’ on A’A and AA’ produced respectively so that 
the following relations may hold simultaneously, 

| Ἐπ τες (α), 
TTA AE AKG ACME τ τ (iS); 
TEA AEA AM 

Analysis. 

Suppose the problem solved and the points MW, H, H’ all 
found. 

Place AK at right angles to 4A’, and draw A’K’ parallel 
and equal to AK. Join KM, K’M, and produce them to meet 
Κ΄ Α΄, KA respectively in £, F. Jom KK’, draw HG through 
E parallel to A’A meeting KF in G, and through M draw QM N 
parallel to AK meeting HG in Q and KK’ in VN. 

Now HAY AM ἐκ AM by (8), 

= FA : AM, by similar triangles, 


whence HA=FA. 
Similarly ἘΠ 11.5.00} 
Next, 


FA+AM: A’K'+A’M=AM: A’'M 
=AK+ AM: EA’ + A’'M, by similar triangles. 


76 ARCHIMEDES 


Therefore 
(FA + AM).(HA'+ A’M)=(KA 4+ AM).(K'A'+ ὀἨΑ4 
Take AR along AH and A’R’ along A’H’ such that 
ATi =A = AK, 
Then, since FA + AM= HM, HA’ + A’M=MBA", we have 
EM ME GM, ΜῊ. τ 2 eee (δ). 
(Thus, if R falls between A and H, FR’ falls on the side of H’ 


remote from A’, and vice versa.) 


Now C: D=HM : MH", by hypothesis, 
=H, MM 
=RM. MR’: MH™, by-(0). 
Measure MV along MN so that MWV=A’M. Join A’V and 
produce it both ways. Draw RP, R’P’ perpendicular to RR’ 
meeting A’V produced in P, P’ respectively. Then, the angle 
MA'V being half a right angle, PP’ is given in position, and, 
since ft, R’ are given, so are P, P’. 
And, by parallels, 
P'V:PV=R'M : MR. 


ON THE SPHERE AND CYLINDER II. ΤΕ] 


Therefore PV.P’V: PV?=RM.MR’: RM’. 

But PY" =2hM*. 

Therefore PV.P’'V=2RM. MR’. 
And it was shown that 

FM Meee = CD. 

Hence PV eevee == 20). 1) 
But MH’ = A'M+ A’E= VM + MQ= QV. 

Therefore QV*: PV.P’V=D: 26, ἃ given ratio. 
Thus, if we take a line p such that 

eae = pis PPS, 

and if we describe an ellipse with PP’ as a diameter and p as 
the corresponding parameter [= DD”®/PP’ in the ordinary 
notation of geometrical conics], and such that the ordinates to 
PP’ are inclined to it at an angle equal to half a right angle, 


i.e. are parallel to QV or AK, then the ellipse will pass 
through ᾧ. 


Hence Q lies on an ellipse given in position. 
Again, since HK is a diagonal of the parallelogram GK’, 
GQ.QN = AA’. A’'K". 


If therefore a rectangular hyperbola be described with AG, 
KK’ as asymptotes and passing through A’, it will also pass 
through Q. 


Hence Q lies on a given rectangular hyperbola. 


Thus @ is determined as the intersection of a given ellipse 


* There is a mistake in the Greek text here which seems to have escaped the 
notice of all the editors up to the present. The words are ἐὰν ἄρα ποιήσωμεν, ws 
τὴν A πρὸς Thy διπλασίαν τῆς 1', οὕτως τὴν TY πρὸς ἄλλην τινὰ ws τὴν ®, i.e. (with 
the lettering above) ‘‘If we take a length p such that ἢ : 26 -- PP’: p.” This 
cannot be right, because we should then have 

QV Be Vere Vi PP Oop: 
whereas the two latter terms should be reversed, the correct property of the 
ellipse being 

OV ΡΥ ΤΡ ΞΕ: PP’ [Apollonius I. 21] 

The mistake would appear to have originated as far back as Eutocius, but I 
think that Eutocius is more likely to have made the slip than Diocles himself, 
because any intelligent mathematician would be more likely to make such a slip 
in writing out another man’s work than to overlook it if made by another. 


78 ARCHIMEDES 


and a given hyperbola, and is therefore given. Thus M is 
given, and H, H’ can at once be found. 


Synthesis. 

Place AA’, AK at right angles, draw A’K’ parallel and 
equal to AK, and join KK’. 

Make AR (measured along A’A produced) and A’R’ 


(measured along AA’ produced) each equal to AK, and 
through R, R’ draw perpendiculars to RR’. 


Then through A’ draw PP’ making an angle (A A’P) with 
AA’ equal to half a right angle and meeting the perpendiculars 
just drawn in P, P’ respectively. 


Take a length p such that 
De — ps Pr. 
and with PP’ as diameter and p as the corresponding parameter 


describe an ellipse such that the ordinates to PP’ are inclined 
to it at an angle equal to AA’P, 1.6. are parallel to AK. 


With asymptotes KA, KK’ draw a rectangular hyperbola 
passing through A’. 

Let the hyperbola and ellipse meet in Q, and from Q draw 
QMVN perpendicular to AA’ meeting AA’ in M, PP’ in V 
and KK'in N. Also draw GQE parallel to A.A’ meeting AK, 
A’K’ respectively in G, £. 

Produce KA, K’M to meet in F’. 

Then, from the property of the hyperbola, 

GORON=AA’ AKG 
and, since these rectangles are equal, KE is a straight line. 


Measure AH along AR equal to AF, and A’H’ along A’R’ 
equal to A’E. 


From the property of the ellipse, 
OV? PV Eve ear 
ΞΞ 20. 


* Here too the Greek text repeats the same error as that noted on p. 77. 


ON THE SPHERE AND CYLINDER II. 79 


And, by parallels, 
PV:P’'V=RM: R’'M, 


or PV PVP Vj RM MR RM, 
while P’V?=2R’M’, since the angle RA’P is half a mght 
angle. 
Therefore PV,P’V=2RM. MR’, 
whence OVA ΜΕ ME =D: 2C. 
But OV Ae AM MT. 


Therefore EM oi MH? = C =D, 
Again, by similar triangles, 
FA+AM:K’'A'+A'M=AM:A'M 
=KA+AM:£EA’'+ A’M. 
Therefore 
(FA + AM).(EA’+A’'M)=(KA+AM).(K'A'+A’'M) 
or HM.MH’=RM.MR’. 
It follows that 
HM .MH’-MH*—=C: D, 
or ἘΠΕ 5! SCR De eee ee ree (a). 
Also FRAG Ann EA - ΑΝ 
= A’K’: Α΄, by similar 
triangles ...(8), 
and HATA EA” AM 


Hence the pomts M, H, H’ satisfy the three given 
relations. ] 


Proposition 5. (Problem.) 


To construct a segment of a sphere similar to one segment 
and equal in volume to another. 

Let ABB’ be one segment whose vertex is A and whose 
base is the circle on BB’ as diameter; and let DEF be another 
segment whose vertex is D and whose base is the circle on HF 


80 ARCHIMEDES 


as diameter. Let AA’, DD’ be diameters of the great circles 
passing through BB’, EF respectively, and let O, C be the 


respective centres of the spheres. 


Suppose it required to draw a segment similar to DEF and 
equal in volume to ABB’, 


Analysis. Suppose the problem solved, and let def be the 
required segment, d being the vertex and ef the diameter of 
the base. Let dd’ be the diameter of the sphere which bisects 
ef at right angles, c the centre of the sphere. 


ZN 
Ne 


Ὁ In 


a’ 


Let M, G,g be the points where BB’, EF, ef are bisected 
at right angles by A.A’, DD’, dd’ respectively, and produce OA, 
CD, cd respectively to H, K, k, so that 


OA'+A’M: A’/M=HM:MA 
CD 2a Ga G= KE GD | : 
οα΄ -αὐσ: αἀὐστεῖρ: gd 
and suppose cones formed with vertices H, K, k and with the 


same bases as the respective segments. The cones will then be 
equal to the segments respectively [Prop. 2]. 


Therefore, by hypothesis, 
the cone HBB’ =the cone kef. 


ON THE SPHERE AND CYLINDER II. 81 


Hence 
(circle on diameter BB’) : (circle on diameter ef)=kg : HM, 
so that 1803) 2502 ἘΞ Τῶν 3 61 SA pa ere (1). 


But, since the segments DEF, def are similar, so are the 
cones KEF, kef: 


Therefore KG: EF=kg: 67: 
And the ratio KG: HF is given. Therefore the ratio kg : ef 


is given. 
Suppose a length & taken such that 


gfe δι παν (2). 
Thus £ is given. 


Again, since kg: HM=BB”: ef’=ef : R, by (1) and (2), 
suppose a length S taken such that 


ef? = BB’ .8, 
or Bierce, -— BB: 8. 
Thus BB sef—er 2 S=S : fh, 


and ef, S are two mean proportionals in continued proportion 
between BB’, R. 


Synthesis. Let ABB’, DEF be great circles, AA’, DD’ 
the diameters bisecting BB’, HF at right angles in M, G 
respectively, and O, C the centres. 

Take H, K in the same way as before, and construct the 
cones HBB’, K EF, which are therefore equal to the respective 
segments ABB’, DEF. 

Let R be a straight line such that 

1G ie — EM: he, 
and between BB’, R take two mean proportionals ef, S. 

On ef as base describe a segment of a circle with vertex d 
and similar to the segment of a circle DHF. Complete the 
circle, and let dd’ be the diameter through d, and ὁ the centre. 
Conceive a sphere constructed of which def is a great circle, 
and through ef draw a plane at right angles to dd’. 


τῆς Ἄν. 6 


82 ARCHIMEDES 


Then shall def be the required segment of a sphere. 


For the segments DEF, def of the spheres are similar, like 
the circular segments DEF, def. 


Produce cd to k so that 
cd’+d'q:d’g=kg : gd. 
The cones KEF, kef are then similar. 
Therefore kg: ef=KG: EF=HM: R, 
whence kg: HM=ef: R. 
But, since BB’, ef, S, R are in continued proportion, 
BB =o -— BD 
=ef: ἢ 
=kg > iM: 
Thus the bases of the cones HBB’, kef are reciprocally 
proportional to their heights. The cones are therefore equal, 


and def is the segment required, being equal in volume to the 
cone kef. [Prop. 2] 


Proposition 6. (Problem.) 


Given two segments of spheres, to find a third segment of a 
sphere similar to one of the given segments and having its 
surface equal to that of the other. 


Let ABB’ be the segment to whose surface the surface of 
the required segment is to be equal, ABA’B’ the great circle 
whose plane cuts the plane of the base of the segment ABB’ at 
right angles in BB’. Let AA’ be the diameter which bisects 
BB’ at right angles. 


Let DEF be the segment to which the required segment 
is to be similar, DED’F the great circle cutting the base of the 
segment at right angles in HF. Let DD’ be the diameter 
bisecting HF at right angles in G. 


Suppose the problem solved, def being a segment similar 
to DEF and having its surface equal to that of ABB’; and 


ON THE SPHERE AND CYLINDER II. 83 


complete the figure for def as for DHF, corresponding points 
being denoted by small and capital letters respectively. 


Ό 


α' 
Join AB, DF, df, 


Now, since the surfaces of the segments def, ABB’ are equal, 
so are the circles on df, AB as diameters ; [I. 42, 43] 


that is, dj — Aap. 
From the similarity of the segments DEF, def we obtain 
d'd:dg=D'D: DG, 


and dg: df= DG: DF; 
whence d'd:df=D'D : DF, 
or d’d: AB=D’'D: DF. 


But AB, D’D, DF are all given; 
therefore d’d is given. 
Accordingly the synthesis is as follows. 
Take d’d such that 
A DEE eusis te saite τοῖοι Cl). 


Describe a circle on d’d as diameter, and conceive a sphere 
constructed of which this circle is a great circle. 


6—2 


84 ARCHIMEDES 


Divide d’d at g so that 
d'g: ga = D'G : GD, 

and draw through g a plane perpendicular to d’d cutting off 
the segment def of the sphere and intersecting the plane of the 
great circle in ef The segments def, DEF are thus similar, 
and dg: df=DG: DF. 

But from above, componendo, 

add. sdg= DD: DG. 

Therefore, ex aequalt, dd:df=D'D: DF, 
whence, by (1), df= AB. 

Therefore the segment def has its surface equal to the 
surface of the segment ABB’ [1. 42, 43], while it is also similar 
to the segment DEF. 


Proposition 7. (Problem.) 

From a gwen sphere to cut off a segment by a plane so that 
the segment may have a given ratio to the cone which has the same 
base as the segment and equal height. 

Let AA’ be the diameter of a great circle of the sphere. 
It is required to draw a plane at right angles to AA’ cutting 
off a segment, as ABB’, such that the segment ABB’ has to 
the cone ABB’ a given ratio. 

Analysis. 

Suppose the problem solved, and let the plane of section 
cut the plane of the great circle in BB’, and the diameter 
AA’ in M. Let O be the centre of the sphere. 


D 


: P | 


Produce OA to H so that _ 
OA’ +A’'’M : A‘M= AM 2 MA... ade: (1). 


ON THE SPHERE AND CYLINDER II. 85 


Thus the cone HBB’ is equal to the segment ABB’. [Prop. 2] 


Therefore the given ratio must be equal to the ratio of the 
cone HBB’ to the cone ABB’, 1.6. to the ratio HM : MA. 


Hence the ratio OA’+ A’M : A’M is given; and therefore 
A’M is given. 


διορισμός. 
Now OA AIM > OAl 3A’ A, 
so that OA’ ΑΜ: A'M>OA'+A'A: A’A 


Ser 5. ΟΣ 


Thus, in order that a solution may be possible, wt is a 
necessary condition that the given ratio must be greater than 
8: 2. 


The synthesis proceeds thus. 


Let AA’ be a diameter of a great circle of the sphere, O the 
centre, 


Take a line DE, and a point F on it, such that DE: EF is 
equal to the given ratio, being greater than 3: 2. 


Now, since ΟΣ ACA Al A= 352: 
DE: EF >OA'+ 444: A'A, 
so that DESEE > OA: ACA. 
Hence a point M can be found on AA’ such that 
DDR pb OAs Ag Mensa cs access ca): 


Through M draw a plane at right angles to AA’ intersecting 
the plane of the great circle in BB’, and cutting off from the 
sphere the segment ABB’. 


As before, take H on OA produced such that 
OA'+ A'M: A’M=HM: MA. 
Therefore HM: MA=DE: EF, by means of (2). 


It follows that the cone HBB’, or the segment ABB’, is to 
the cone ABB’ in the given ratio DE: EF. 


86 ARCHIMEDES 


Proposition 8. 


If a sphere be cut by a plane not passing through the centre 
into two segments A’BB’, ABB’, of which A’BB’ is the greater, 
then the ratio 
(segmt. A’BB’) : (segmt. ABB’) 

< (surface of A’BB’)’ : (surface of ABB’y 
but > (surface of A’BB’): : (surface of ABB’)*. 

Let the plane of section cut a great circle A’BAB’ at right 


angles in BB’, and let AA’ be the diameter bisecting BB’ at 
right angles in M. 


Let O be the centre of the sphere. 
Join A’B, AB. 


As usual, take H on OA produced, and H’ on OA’ produced, 
so that 
OAs AM: ACM = πιο τ  νΝ 


0OA+AM:AM=H'M:MA’............. (2), 


and conceive cones drawn each with the same base as the two 
segments and with apices H, H’ respectively. The cones are 
then respectively equal to the segments [Prop. 2], and they 
are in the ratio of their heights HM, H’M. 

Also 
(surface of A’ BB’): (surface of ABB’)= A’B?: AB? Π]. 42, 43] 

-- 4 :- ΑΨ. 

* This is expressed in Archimedes’ phrase by saying that the greater seg- 

ment has to the lesser a ratio ‘‘less than the duplicate (διπλάσιον) of that which 


the surface of the greater segment has to the surface of the lesser, but greater 
than the sesquialterate (ἡμιόλιον) [of that ratio].” 


ON THE SPHERE AND CYLINDER II. 87 


We have therefore to prove 
(a) that HOM = ME 7 : MA 
(b) that H’M: MH > A'M?: MA}. 

(a) From (2) above, 

ΑΜ: AM=H'M:0A+AM 
= H'A’: OA’, since OA =OA’. 

Since A’M> AM, H’A'>OA’; therefore, if we take K on 
H’A’ so that OA’=A’K, Καὶ will fall between H’ and A’. 

And, by (1), PAM 4.77. Ke 2 Mid. 


Thus ΚΗ: ML = Π΄ Δ’: ATK since AK = OA. 
> H’M : MK. 
Therefore A’/M.MH < KM”. 


It follows that 
EM SN MT Ms ΜΠΗ͂Ι 
or FEM MH ΞΜ’. ΜΗ: 
< AMS eA by i): 
(oO) simee OA SOA, 
AMM = -A'O OA. 
or AME OA AVA > AM 
< H'A’: A’'M, by means of (2). 
Therefore AIM <A’ OAs 
ΞΘ ΠΕ 
Take a point V on 4΄4 such that 
eA AL A Ke 
Thus EVGA ESPA CHG Ξ AN Cie ACen τ: (3). 
Also PTAA NAN 2 aA EK 
and, componendo, 
TONG Al N= NK =A Ke 
whence AGNG Ale = HNC ΝΣ 


88 ARCHIMEDES 
Therefore, by (3), 
H'A’: A'K =H'N’: NR’. 


Now ΗΜ: MK>H'N: NK. 
Therefore A’M’?: MK*>H'A’': A’K 
SH AOA’ 


> A’M: MA, by (2), as above, 
> OA’+A’M : MH, by (1), 
>KM: MH. 
Hence AM? ΜΗ ΞΟΠ: MK). (Bee 
> (KM : MH). (KM: MA"). 
It follows that 
H'M : MH > KM’: MH? 
> A’M? : AM}, by (1). 


[The text of Archimedes adds an alternative proof of this 
proposition, which is here omitted because it is in fact neither 
clearer nor shorter than the above. | 


Proposition 9. 


Of all segments of spheres which have equal surfaces the 
henusphere is the greatest in volume. 


Let ABA’B’ be a great circle of a sphere, AA’ being 
a diameter, and O the centre. Let the sphere be cut by 
a plane, not passing through O, perpendicular to AA’ (at JZ), 
and intersecting the plane of the great circle in BB’. The 
segment ABB’ may then be either less than a hemisphere as 
in Fig. 1, or greater than a hemisphere as in Fig. 2. 


Let DED’E’ be a great circle of another sphere, DD’ 
being a diameter and C the centre. Let the sphere be cut by 
a plane through Οὐ perpendicular to DD’ and intersecting the 
plane of the great circle in the diameter EE’. 


ON THE SPHERE AND CYLINDER II. 89 


Suppose the surfaces of the segment ABB’ and of the 
hemisphere DEE’ to be equal. 


Since the surfaces are equal, AB = DE. [I. 42, 43] 
Now, in Fig. 1, ABS 247 and <2A07, 
and, in Fig. 2, AB*<2AM* and >2A0", 
Hence, if R be taken on AA’ such that 
AR? =4AB?, 
R will fall between O and MM. 
Also, since AB?= DE’, AR=CD. 


Produce OA’ to K so that OA’ = A’K, and produce A’A to 
HT so that 
AUTG SA ELA ACM. 
or, componendo, A’K+A’M: A’M=HM: MA............ (1). 
Thus the cone HBB’ is equal to the segment ABB’. 
[Prop. 2] 


Again, produce CD to F so that CD= DF, and the cone 
FEE’ will be equal to the hemisphere DEE’. [Prop. 2] 


Now AkR.RA'’>AM.MA’, 
and AR? =4AB*=4AM. AA’=AM.A’K. 


90 ARCHIMEDES 


Hence 
ARh.RA'+RA*>AM.MA’+ AM.A’K, 
or AA’. AR>AM.MK 
> HM. A’M, by (1). 
Therefore AA’: A’M>HM: AR, 


or AB*: BMP > HM > AR, 
1.e. AR’: BM*>HM: 2AR, since AB? =2A R?, 
> HM: CF. 


Thus, since AR = CD, or CH, 
(circle on diam. FE’) : (circle on diam. BB’)> HM: CF. 
It follows that 

(the cone FEE’) > (the cone HBB’), 


and therefore the hemisphere DHL’ is greater in volume than 
the segment ABB". 


MEASUREMENT OF A CIRCLE. 


Proposition 1. 


The area of any circle is equal to a right-angled triangle in 
which one of the sides about the right angle is equal to the radius, 
and the other to the circumference, of the circle. 


Let ABCD be the given circle, K the triangle described. 


Then, if the circle is not equal to Κ΄, it must be either 
greater or less. 


I. If possible, let the circle be greater than K. 

Inscribe a square ABCD, bisect the arcs AB, BC, CD, DA, 
then bisect (if necessary) the halves, and so on, until the sides 
of the inscribed polygon whose angular points are the points of 
division subtend segments whose sum is less than the excess of 
the area of the circle over K. 


92 ARCHIMEDES 


Thus the area of the polygon is greater than ζ΄. 


Let AZ be any side of it, and ON the perpendicular on AZ 
from the centre O. 


Then ON is less than the radius of the circle and therefore 
less than one of the sides about the right angle in K. Also the 
perimeter of the polygon is less than the circumference of the 
circle, i.e. less than the other side about the right angle in K. 


Therefore the area of the polygon is less than K; which is 
inconsistent with the hypothesis. 


Thus the area of the circle is not greater than K. 


II. If possible, let the circle be less than K. 


Circumscribe a square, and let two adjacent sides, touching 
the circle in HZ, H, meet in 7. Bisect the arcs between adjacent 
points of contact and draw the tangents at the points of 
bisection. Let A be the middle point of the are HH, and FAG 
the tangent at A. 


Then the angle TAG is a right angle. 
Therefore LG> GA 
> GH. 


It follows that the triangle /7G is greater than half the area 
TEAH. 


Similarly, if the are AH be bisected and the tangent at the 
point of bisection be drawn, it will cut off from the area GAH 
more than one-half. 


Thus, by continuing the process, we shall ultimately arrive 
at a circumscribed polygon such that the spaces intercepted 
between it and the circle are together less than the excess of 
K over the area of the circle. 


Thus the area of the polygon will be less than K. 


Now, since the perpendicular from O on any side of the 
polygon is equal to the radius of the circle, while the perimeter 
of the polygon is greater than the circumference of the circle, 
it follows that the area of the polygon is greater than the 
triangle K; which is impossible. 


MEASUREMENT OF A CIRCLE. 93 


Therefore the area of the circle is not less than K. 


Since then the area of the circle is neither greater nor less 
than K, it is equal to it. 


Proposition 2. 


The area of a circle 15 to the square on its diameter as 11 
to 14. 


[The text of this proposition is not satisfactory, and Archi- 
medes cannot have placed it before Proposition 3, as the 
approximation depends upon the result of that proposition,] 


Proposition S. 


The ratio of the circumference of any circle to its diameter 
as less than 31 but greater than 319. 


[In view of the interesting questions arising out of the 
arithmetical content of this proposition of Archimedes, it is 
necessary, in reproducing it, to distinguish carefully the actual 
steps set out in the text as we have it from the intermediate 
steps (mostly supplied by Eutocius) which it is convenient to 
put in for the purpose of making the proof easier to follow. 
Accordingly all the steps not actually appearing in the text 
have been enclosed in square brackets, in order that it may be 
clearly seen how far Archimedes omits actual calculations and 
only gives results. It will be observed that he gives two 
fractional approximations to 4/3 (one being less and the other 
greater than the real value) without any explanation as to how 
he arrived at them; and in like manner approximations to the 
square roots of several large numbers which are not complete 
squares are merely stated. These various approximations and 
the machinery of Greek arithmetic in general will be found 
discussed in the Introduction, Chapter IV.] 


I, Let AB be the diameter of any circle, O its centre, AC 
the tangent at A; and let the angle AOC be one-third of a 
right angle. 


Ὄπ ΠῚ 


94 ARCHIMEDES 


Then OA: AC [2/821] > 265: 153... ἘΠ @), 
and OC: CAl=]2 21] =306 +153...) eee (2). 
First, draw OD bisecting the angle AOC and meeting AC 
in D. 
Now COS OAS CD=DA, [Eucl. VI. 3] 
so that [CO+ 0A :0A=CA: DA, or] 
CO+0A =CA=O0A : AD. 
Therefore [by (1) and (2)] 
OA ADS O01 2 108. vs oe... sca eee (3). 
Hence OD* = AD |. OAS + AD): AD? 
> (571? + 1685) : 1537] 
> 349450 : 23409, 
so that OD DAS 0 ΘΠ (4). 


Secondly, let OF bisect the angle AOD, meeting AD in £. 
[Then DO: 0A=DE EA, 
so that DO+0A:DA=0A: AE] 
Therefore OA : AE[>(5911 +571) : 153, by (3) and (4)] 
ἈΠ ΡΠ (5). 


MEASUREMENT OF A CIRCLE. 95 


[It follows that 
OE? : HA? > {(11624)? + 1537} : 153? 
> (135053438 + 23409) : 23409 
> 137394338 : 23409. ] 
Thus OR TAS ULI Zhe LSS. ere iced cnt oles cotint (6). 


Thirdly, let OF bisect the angle AO# and meet AF in F. 


We thus obtain the result [corresponding to (8) and (5) 
above] that 
OA : AF[> (11623 + 11722) : 153] 


ΣΣ 299 159. 2 tnne saneacetee Gi): 
[Therefore OF? : ΓΑ" {(23844) + 1537} : 153? 

> 54721321, : 23409.] 
Thus OT Pe 2350 ον sacisusesionee (8). 


Fourthly, let OG bisect the angle AOF, meeting AF in G. 
We have then 
OA : AG [> (23344 + 23394) : 153, by means of (7) and (8)] 
> 46732 : 158. 


Now the angle AOC, which is one-third of a right angle, 
has been bisected four times, and it follows that 


ZAOG = x; (a right angle). 


Make the angle 4 ΟΠ on the other side of OA equal to the 
angle AOG, and let GA produced meet OH in H. 
Then Z GOH = x, (a right angle), 


Thus GH is one side of a regular polygon of 96 sides cir- 
cumscribed to the given circle. 


And, since OA : AG > 46736 : 153, 
while AB=20A, GH=2AG, 
it follows that 
AB : (perimeter of polygon of 96 sides) [> 46784 : 153 x 96] 
> 46734 : 14688. 


96 ARCHIMEDES 


14688 _, | 6674 
467384 ° ' 46733 


6673 
|< al ΕἸ 


But 


<3. 

Therefore the circumference of the circle (being less than 
the perimeter of the polygon) is a fortiori less than 3} times 
the diameter AB. 

II. Next let AB be the diameter of a circle, and let AC, 
meeting the circle in Οὐ, make the angle CAB equal to one-third 
of aright angle. Join BC. 

Then AC: CBlH V3 21) < 13851 : 780. 

First, let AD bisect the angle BAC and meet BC in d and 
the circle in D. Join BD. 

Then ZBAD=ZdAC 

= ZdBD, 
and the angles at D, C are both right angles. 
It follows that the triangles A DB, [ACd], BDd are similar. 


O aes 


Therefore AD DB BD Dd 
[= AC: Cd] 
=AB: Bd [Eucl. VI. 3] 
=AB+ AC: Bd+Cd 
=AB+AC: BC 


or BA+AC: BC=AD: DB. 


MEASUREMENT OF A CIRCLE. 97 


[But AC : CB< 1351 : 780, from above, 
while BA: BC=2:1 
= 1560 : 780.] 
Therefore Fes) ΞΘ: | ener eee (1). 
[Hence AB’: BD’ «(2911 + 780") : 780° 
< 9082321 : 608400.] 
Thus CANE eo) AAU SS GO uta ares ace (2). 


Secondly, let AE bisect the angle BAD, meeting the circle 
in #; and let BE# be joined. 


Then we prove, in the same way as before, that 
AE: EB[=BA+AD: BD 
< (30132 + 2911) : 780, by (1) and (2)] 
< 59242 : 780 
< 59242 x 4: 780 x 4 
SLO): AAO iia oto we ebm eccnane sees (3). 
[Hence AB’: ΒΕ" < (1823? + 240°) : 240? 
< 3880929 : 57600.] 

Therefore AB: BE < 18382, θα: τ τ (4). 
Thirdly, ἰοὺ AF bisect the angle BAL, meeting the circle 
ἴῃ 

Thus AF: FB[=BA+AE: BE 

< 3661-2 : 240, by (3) and (4)] 

< 36614, x 45: 240 χ τῷ 

τ Τ0 0 7 1G Orceson. acceceewecssr.tee(O)s 
[1t follows that 


AB* : BF’ < (1007? + 66’) : 66° 
< 1018405 : 4356.] 
Therefore AT BP οτος. τ... (6). 


Fourthly, let the angle BAF be bisected by AG meeting the 
circle in G. 


Then AG:GB[=BA+ AF: BF] 
< 20164 : 66, by (5) and (6). 


98 ARCHIMEDES 


[And AB’: BG’ < {(20161)? + 66 : 66° 
< 40692841. : 4356.] 
Therefore AB : BG < 20174 : 66, 
whence BG: ABS 66 220114: «000.0005, eee UG) 


[Now the angle BAG which is the result of the fourth bisection 
of the angle BAC, or of one-third of a right angle, is equal to 
one-fortyeighth of a right angle. 


Thus the angle subtended by BG at the centre is 
a (a right angle).] 
Therefore BG is a side of a regular inscribed polygon of 96 
sides. 
It follows from (7) that 
(perimeter of polygon) : AB [> 96 x 66 : 20171 
> 6336 : 20174. 
And a > BH. 
as 
Much more then is the circumference of the circle greater than 
34° times the diameter. 
Thus the ratio of the circumference to the diameter 


< 34 but > 312. 


ON CONOIDS AND SPHEROIDS. 


Introduction*. 


‘‘ ARCHIMEDES to Dositheus greeting. 


In this book I have set forth and send you the proofs of the 
remaining theorems not included in what I sent you before, and 
also of some others discovered later which, though I had often 
tried to investigate them previously, I had failed to arrive at 
because I found their discovery attended with some difficulty. 
And this is why even the propositions themselves were not 
published with the rest. But afterwards, when I had studied 
them with greater care, I discovered what I had failed in 
before. 


Now the remainder of the earlier theorems were propositions 
concerning the right-angled conoid [paraboloid of revolution] ; 
but the discoveries which I have now added relate to an obtuse- 
angled conoid [hyperboloid of revolution] and to spheroidal 
figures, some of which I call oblong (παραμάκεα) and others flat 
(ἐπιπλατέα). 


I. Concerning the right-angled conoid it was laid down 
that, if a section of a right-angled cone [a parabola] be made to 
revolve about the diameter [axis] which remains fixed and 


* The whole of this introductory matter, including the definitions, is trans- 
lated literally from the Greek text in order that the terminology of Archimedes 
may be faithfully represented. When this has once been set out, nothing will 
be lost by returning to modern phraseology and notation. These will accordingly 
be employed, as usual, when we come to the actual propositions of the treatise. 


7—2 


100 ARCHIMEDES 


return to the position from which it started, the figure compre- 
hended by the section of the right-angled cone is called a right- 
angled conoid, and the diameter which has remained fixed 
is called its axis, while its vertex is the point in which the 
axis meets (ἅπτεται) the surface of the conoid. And if a plane 
touch the right-angled conoid, and another plane drawn parallel 
to the tangent plane cut off a segment of the conoid, the base 
of the segment cut off is defined as the portion intercepted by 
the section of the conoid on the cutting plane, the vertex 
[of the segment] as the point in which the first plane touches 
the conoid, and the axis [of the segment] as the portion cut 
off within the segment from the line drawn through the vertex 
of the segment parallel to the axis of the conoid. 


The questions propounded for consideration were 


(1) why, if a segment of the right-angled conoid be cut off 
by a plane at right angles to the axis, will the segment so cut 
off be half as large again as the cone which has the same base 
as the segment and the same axis, and 


(2) why, if two segments be cut off from the right-angled 
conoid by planes drawn in any manner, will the segments so cut 
off have to one another the duplicate ratio of their axes. 


II. Respecting the obtuse-angled conoid we lay down the 
following premisses. If there be in a plane a section of an 
obtuse-angled cone [a hyperbola], its diameter [axis], and the 
nearest lines to the section of the obtuse-angled cone [1.6. the 
asymptotes of the hyperbola], and if, the diameter [axis] 
remaining fixed, the plane containing the aforesaid lines be 
made to revolve about it and return to the position from which 
it started, the nearest lines to the section of the obtuse-angled 
cone [the asymptotes] will clearly comprehend an isosceles cone 
whose vertex will be the point of concourse of the nearest lines 
and whose axis will be the diameter [axis] which has remained 
fixed. The figure comprehended by the section of the obtuse- 
angled cone is called an obtuse-angled conoid [hyperboloid of 
revolution], its axis is the diameter which has remained fixed, 
and its vertex the point in which the axis meets the surface 


ON CONOIDS AND SPHEROIDS. 101 


of the conoid. The cone comprehended by the nearest lines to 
the section of the obtuse-angled cone is called [the cone] 
enveloping the conoid (περιέχων τὸ κωνοειδές), and the 
straight line between the vertex of the conoid and the vertex 
of the cone enveloping the conoid is called [the line] adjacent 
to the axis (ποτεοῦσα τῷ ἄξονι). And if a plane touch the 
obtuse-angled conoid, and another plane drawn parallel to the 
tangent plane cut off a segment of the conoid, the base of 
the segment so cut off is defined as the portion intercepted by 
the section of the conoid on the cutting plane, the vertex [of 
the segment] as the point of contact of the plane which touches 
the conoid, the axis [of the segment] as the portion cut off 
within the segment from the line drawn through the vertex of 
the segment and the vertex of the cone enveloping the conoid ; 
and the straight line between the said vertices is called 
adjacent to the axis. 


Right-angled conoids are all similar; but of obtuse-angled 
conoids let those be called similar in which the cones enveloping 
the conoids are similar. 


The following questions are propounded for consideration, 


(1) why, if a segment be cut off from the obtuse-angled 
conoid by a plane at right angles to the axis, the segment so 
cut off has to the cone which has the same base as the segment 
and the same axis the ratio which the line equal to the sum 
of the axis of the segment and three times the line adjacent 
to the axis bears to the line equal to the sum of the axis of 
the segment and twice the line adjacent to the axis, and 


(2) why, if a segment of the obtuse-angled conoid be cut 
off by a plane not at right angles to the axis, the segment so 
cut off will bear to the figure which has the same base as 
the segment and the same axis, being a segment of a cone* 
(ἀπότμαμα κώνου), the ratio which the line equal to the sum 
of the axis of the segment and three times the line adjacent 
to the axis bears to the line equal to the sum of the axis of the 
segment and twice the line adjacent to the axis. 


* A segment of a cone is defined later (p. 104). 


102 ARCHIMEDES 


III. Concerning spheroidal figures we lay down the follow- 
ing premisses. Ifa section of an acute-angled cone [ellipse] be 
made to revolve about the greater diameter [major axis] which 
remains fixed and return to the position from which it started, 
the figure comprehended by the section of the acute-angled 
cone is called an oblong spheroid (παραμᾶκες σφαιροειδές). 
But if the section of the acute-angled cone revolve about the 
lesser diameter [minor axis] which remains fixed and return 
to the position from which it started, the figure comprehended 
by the section of the acute-angled cone is called a flat spheroid 
(ἐπιπλατὺ σφαιροειδές). In either of the spheroids the axis 
is defined as the diameter [axis] which has remained fixed, the 
vertex as the point in which the axis meets the surface of the 
spheroid, the centre as the middle point of the axis, and the 
diameter as the line drawn through the centre at right angles 
to the axis. And, if parallel planes touch, without cutting, 
either of the spheroidal figures, and if another plane be drawn 
parallel to the tangent planes and cutting the spheroid, the 
base of the resulting segments is defined as the portion inter- 
cepted by the section of the spheroid on the cutting plane, their 
vertices as the points in which the parallel planes touch the 
spheroid, and their axes as the portions cut off within the 
segments from the straight line joining their vertices. And 
that the planes touching the spheroid meet its surface at one 
point only, and that the straight line joining the points of 
contact passes through the centre of the spheroid, we shall 
prove. Those spheroidal figures are called similar in which 
the axes have the same ratio to the ‘diameters. And let 
segments of spheroidal figures and conoids be called similar if 
they are cut off from similar figures and have their bases 
similar, while their axes, being either at right angles to the 
planes of the bases or making equal angles with the corre- 
sponding diameters [axes] of the bases, have the same ratio 
to one another as the corresponding diameters [axes] of the 
bases. 

The following questions about spheroids are propounded for 
consideration, 

(1) why, if one of the spheroidal figures be cut by a plane 


ON CONOIDS AND SPHEROIDS. 103 


through the centre at right angles to the axis, each of the 
resulting segments will be double of the cone having the same 
base as the segment and the same axis; while, if the plane of 
section be at right angles to the axis without passing through 
the centre, (a) the greater of the resultmg segments will bear 
to the cone which has the same base as the segment and the 
same axis the ratio which the line equal to the sum of half the 
straight line which is the axis of the spheroid and the axis of 
the lesser segment bears to the axis of the lesser segment, and 
(b) the lesser segment bears to the cone which has the same 
base as the segment and the same axis the ratio which the line 
equal to the sum of half the straight line which is the axis 
of the spheroid and the axis of the greater segment bears to the 
axis of the greater segment ; 

(2) why, if one of the spheroids be cut by a plane passing 
through the centre but not at right angles to the axis, each of 
the resulting segments will be double of the figure having the 
same base as the segment and the same axis and consisting of a 
segment of a cone*, 

(3) But, if the plane cutting the spheroid be neither 
through the centre nor at right angles to the axis, (a) the 
greater of the resulting segments will have to the figure 
which has the same base as the segment and the same axis 
the ratio which the line equal to the sum of half the line 
joiming the vertices of the segments and the axis of the lesser 
segment bears to the axis of the lesser segment, and (b) the 
lesser segment will have to the figure with the same base 
as the segment and the same axis the ratio which the line 
equal to the sum of half the line joining the vertices of the 
segments and the axis of the greater segment bears to the axis 
of the greater segment. And the figure referred to is in these 
cases also a segment of a cone*. 


When the aforesaid theorems are proved, there are dis- 
covered by means of them many theorems and problems. 

Such, for example, are the theorems 

(1) that similar spheroids and similar segments both of 


* See the definition of a segment of a cone (ἀπότμαμα κώνου) on p. 104. 


104 ARCHIMEDES 


spheroidal figures and conoids have to one another the triplicate 
ratio of their axes, and 

(2) that in equal spheroidal figures the squares on the 

‘diameters’ are reciprocally proportional to the axes, and, if in 

spheroidal figures the squares on the ‘diameters’ are reciprocally 
proportional to the axes, the spheroids are equal. 

Such also is the problem, From a given spheroidal figure 
or conoid to cut off a segment by a plane drawn parallel to a 
given plane so that the segment cut off is equal to a given cone 
or cylinder or to a given sphere. 


After prefixing therefore the theorems and directions (é7u- 
τάγματα) which are necessary for the proof of them, I will 
then proceed to expound the propositions themselves to you. 
Farewell. | 


DEFINITIONS. 

If a cone be cut by a plane meeting all the sides [generators] 
of the cone, the section will be either a circle or a section of an 
acute-angled cone [an ellipse]. If then the section be a circle, 
it is clear that the segment cut off from the cone towards the 
same parts as the vertex of the cone will be a cone. But, if 
the section be a section of an acute-angled cone [an ellipse], let 
the figure cut off from the cone towards the same parts as the 
vertex of the cone be called a segment of a cone. Let the 
base of the segment be defined as the plane comprehended by 
the section of the acute-angled cone, its vertex as the point 
which is also the vertex of the cone, and its axis as the straight 
line joming the vertex of the cone to the centre of the section 
of the acute-angled cone. 


And if a cylinder be cut by two parallel planes meeting all 
the sides [generators] of the cylinder, the sections will be either 
circles or sections of acute-angled cones [ellipses] equal and 
similar to one another. If then the sections be circles, it is 
clear that the figure cut off from the cylinder between the 
parallel planes will be a cylinder. But, if the sections be 
sections of acute-angled cones [ellipses], let the figure cut off 
from the cylinder between the parallel planes be called a 
frustum (τόμος) of a cylinder. And let the bases of the 


ON CONOIDS AND SPHEROIDS. 105 


frustum be defined as the planes comprehended by the sections 
of the acute-angled cones [ellipses], and the axis as the straight 
line joining the centres of the sections of the acute-angled 
cones, so that the axis will be in the same straight line with 
the axis of the cylinder.” 


Lemma. 


Tf in an ascending arithmetical progression consisting of the 
magnitudes A,, As, ... An the common difference be equal to the 
least term A,, then 

n.Ayn<2(A,+A.4+...+An), 
and >2(A,+ 4.-...-- An). 


[The proof of this is given incidentally in the treatise On 
Spirals, Prop. 11. By placing lines side by side to represent 
the terms of the progression and then producing each so as to 
make it equal to the greatest term, Archimedes gives the 
equivalent of the following proof. 


If Sa = 4A, 4 Age... tA, +A,, 
we have also S,= Ant AnitAne2t--- +A. 
And AA eA eggs tars ΞῊ 

Therefore 25, =e yA. 
whence pes Une ots 
and Rig Ape Ohi. 

Thus, if the progression is a, 2a,... na, 

ΓΞ n = ΠῚ ys 
and na < 28, 
but > 28,225] 


Proposition 1. 
WTA, By Ci yey Gnd ΟΝ be 1 Series oF 
magnitudes such that 


Ἢ B= A,B. 
Cij=B, Ὁ and sovon 


106 ARCHIMEDES 


and if A;, Bs, Cs, ...K; and A,, B,, Οἱ, ...K, be two other series 
such that 


B,: B; = B,:B,, and so on 
then (A,+B,+C,+...+ 4) :(A; + B+ Os+... + Κῷ 
=(A,+ B,+C,+...+K.):(As+ Bit... + Ky). 


The proof is as follows. 


ΕΠ Ae 


Since ΕΠ ΞΞῊ 

and bre 2 Ds. 

while 5 5.8). — Bt hg, 

we have, ex aequalt, PARE ΞΘ ΡΣ ᾿ in ay 
Similarly B.C, = B, °C, and’ so on 


Again, it follows from equations (a) that 
AGA ees — 10, Cs Sake 
Therefore 
A,: A,=(A,+ B,+C,4+... +): (4.-- Bo+... + Ko), 
or (A, + B,+0,+...+K)):A,=(4,+ B+O,4+... + Ke): As; 
and Ay A =A AG. 
while from equations (ry) it follows in like manner that 
A;:(A,+Be+O,+...+ K3)=Ag:(As+ Bet Cy +... + Ky). 
By the last three equations, ex aequali, 
(A,+B,+0,4+...+4):(A,+ B+ O,4+...+ Ks) 
=(A,+B,4+ C,4+...+ K,):(A,+ Bit CO, +... + Ky). 
Cor. If any terms in the third and fourth series corre- 
sponding to terms in the first and second be left out, the 


result is the same. For example, if the last terms K;, K, are 
absent, 


(A, +B, +C,+...+K,):(As;+ B+ 034+... + Js) 
=(A,+ B,+C, +... + K.):(Ay+ By + C,4+...+1,), 


where 7 immediately precedes A in each series. 


ON CONOIDS AND SPHEROIDS. 107 


Lemma to Proposition 2. 
[On Spirals, Prop. 10.] 


Tf A,, A;, A;, ...A, be n lines forming an ascending 
arithmetical progression in which the common difference is equal 
to the least term A,, then 


(n+1)A,?4+ A,(A, + A,+ Az +... +An) 
=3(A74+ 4274+ A?+...+A,?). 


ae STEN 


Αι Ao An_-3An-2An-1 


oe 6 6 6. Ὁ 


An AniA,.-2 Ags Ao ΑἹ 


ecwthe ilmes Al, A, 4, An, .-.4)/be placed in ἃ row 
from left to right. Produce A,_,, An», ...A, until they are 
each equal to A,, so that the parts produced are respectively 
equal-to Ἢ A.; ...An.. 


Taking each line successively, we have 
DeAnna Any. 
Cag ian) — oy ΤΑ 2A A, 
(A, + An.) = A,’ + A’. + 2A,.An2, 


See eee eee esse ees eee es eeeeeseeeseseseeserereeeses 


(Ane ate Ay = Ava ἘΝ A; gle ΤΙ Εν : A. 


108 ARCHIMEDES 


And, by addition, 
(n+ 1)A,?=2(AP+A2H+... + An’) 
+2A,.Ay,+2A,.Ano +... ἜΘΗ Ἢ 
Therefore, in order to obtain the required result, we have to 
prove that 
2(A,.An +Ay. Anot...-+An A:)+A(A,+A,+ Ast... +An) 
Ξ As + vce + τ Ὁ (a). 
Now 22,.24., ΞΞΗ 44, because Aj= 247. 
2A, Ay = A.6A,-5, because A,=3A,, 


eee eter ee eeee ere eeesesseeee 


2A — neo — 1 AR 
It follows that 
2(A,.An1+A,.Anot...tAn..A,)+4A,(4,+ A+... +An) 
= A,{A,+8An.+5Anot...+(2n—1)Aj}. 
And this last expression can be proved to be equal to 
Ale Ag ect Ay 
Hor  2-Ξ- Ἢ (ἢ ΤΠ) 
= A,{A,+(n—1)A,} 
= A,{An+2(AniatAnot...+A,)}, 
because (n —1) Ay, = An, + 4, 
+ An.+ A, 


as A, Ἔ τις 
Similarly A* = Al(A, 4 2(A, 4 A, eee 


A,’ = A,(A, + 24,), 
AG SA Ane 
whence, by addition, 
AP+A?l+AP+...+An’ 
= A,{Ayn+3An1 + 5An. t+... +(2n—1)Aj}. 


ON CONOIDS AND SPHEROIDS. 109 


Thus the equation marked (a) above is true; and it follows 
that 


(n+1)A,?+A,(A, +A. + Ast... +4n)=3(Al+ A? +...+A,’). 
Cor. 1. From this it 1s evident that 
i Ages OU Aa 4. Ἔ... Ὁ Ag) canaicteswenies (1). 
Also An =A,{An + 2(Anr+ Ant -.. +A,)}, as above, 
so that A,’ >A,(AntAnat-..+A)), 
and therefore 
A,?+A,(A,+ 4.4+...+An)< 2Ay’. 
It follows from the proposition that 
ΣΉ Ξ ΙΣΤ Ν eee Aly 4) eons cacececvcndees ὯΝ 


Cor. 2. All these results will hold if we substitute similar 


figures for squares on all the lines; for similar figures are in the 
duplicate ratio of their sides. 


[In the above proposition the symbols A,, A,, ...A» have 
been used instead of a, 2a, 3a, ...na in order to exhibit the 
geometrical character of the proof; but, if we now substitute 
the latter terms in the various results, we have (1) 


(n+1)n’?a?+a(a+2a+...+na) 
=3 {a’+ (2a)? + (Ba) +... + (na)’}. 
Therefore αὐ + (2a) + (8a)? +... + (nay 


= : ἰωεἰλλ EO 


ἘΠ n(n +1)(2n - 1) 
=a’, δ ; 


Also (2) <3 (17+ 27+ 37+... +n’), 


and (3) Oy soe (Clea oat ee ear NG 


110 ARCHIMEDES 


Proposition 2. 


If A,, Ay... An be any number of areas such that* 
A, τ αὐ -Ὁ αἱ", 
A,=a.2x + (25) 
A,=a.32 + (32)’, 


A,=a.nx+(nz)’, 


then ne Ant (Art Ayton + An) «(ὦ na): (5 +), 


and n.Ay:(A,+A,+...+An4) > (at nz): (5+) 


For, by the Lemma immediately preceding Prop. 1, 
n.ang< 2(ax+a.2v+...+a.n2), 
and >2(ae+a.24+...+a.n—12). 
Also, by the Lemma preceding this proposition, 
n. (na) « 8 {x + (2x)? + (Bx) + ... + (na)’} 
and or ny... (od Dae 
Hence 
= Pe a < [(aw + x") + {a. 274 (2x)}+...+ {a. ne +(nx)}], 
and 
> [(av + a) + {a. 2. + (22)}+...4+{a.n—-la+(n—12)}], 
an’x n(n) 


2 ie on. sees page wil 


and >A, +Aot+...+ Ay. 
It follows that 


2 2 
n.An:(A,+ Ag+... + An) <n {a. na +(nx)} : er, 


or 


2 3 
or αν. Ant (Ait Ast... Ay) <(a+ne): (54%); 
also me An: (A+ Ast ot Ana) > (ane): (5+), 


* The phraseology of Archimedes here is that associated with the traditional 
method of application of areas: εἴ κα.. παρ᾽ éxdoray αὐτᾶν παραπέσῃ τι χωρίον 
ὑπερβάλλον εἴδει τετραγώνῳ, “if to each of the lines there be applied a space 
[rectangle] exceeding by a square figure.” Thus 4, is a rectangle of height « ap- 
plied to a line a but overlapping it so that the base extends a distance « beyond a. 


ON CONOIDS AND SPHEROIDS. 111 


Proposition 8. 


(1) 1} TP, TP’ be two tangents to any conic meeting in T, 
and if Qq, Q’q be any two chords parallel respectively to TP, 
TP’ and meeting in O, then 


Ὁ: OO 0g =TP*: TP*. 
“ And this is proved in the elements of conics*.” 


(2) If QQ’ bea chord of a parabola bisected in V by the 
diameter PV, and if PV be of constant length, then the areas of 
the triangle PQQ’ and of the segment PQQ’ are both constant 
whatever be the direction of QQ’. 


Let ABB’ be the particular segment of the parabola whose 
vertex is A, so that BB’ is bisected perpendicularly by the axis 
at the point H, where AH = PV. 


Draw QD perpendicular to PV. 


Let pq be the parameter of the principal ordinates, and let 
p be another line of such length that 


OV OD — pin, 
it will then follow that p is equal to the parameter of the ordi- 
nates to the diameter PV, i.e. those which are parallel to QV. 


* i.e. in the treatises on conics by Aristaeus and Euclid. 


112 ARCHIMEDES 


“ For this is proved in the conics*.” 


Thus QV? =p.PV. 

And BH’=p,.AH, while AH=PYV. 
Therefore QV?: BH’ =p: pa. 

But QV?: QD*=p: pa; 
hence rei 1). 

Thus aad — DP V, 
and therefore Mana — ΕΚ)“: 


that is, the area of the triangle PQQ’ is constant so long as PV 
is of constant length. 


Hence also the area of the segment PQQ’ is constant under 
the same conditions; for the segment is equal to 4A PQQ’. 
[Quadrature of the Parabola, Prop. 17 or 24.] 


* The theorem which is here assumed by Archimedes as known can be 
proved in various ways. 

(1) It is easily deduced from Apollonius I. 49 (cf. Apollonius of Perga, 
pp. liii, 39). If in the figure the tangents at A and P be drawn, the former 
meeting PV in E, and the latter meeting the axis in 7, and if AH, PT meet 
at C, the proposition of Apollonius is to the effect that 

CIP 3 Sin ΧΡΗ͂Ν 
where p is the parameter of the ordinates to PV. 
(2) It may be proved independently as follows, 
Let QQ’ meet the axis in O, and let QM, Q’M’, PN be ordinates to the axis. 


Then AM: AM'=QM?: Q'M?=0M?: 0M", 
whence AM :MM’=0OM?: OM?- OM” 
=0OM?:(OM-OM’). MM’, 
so that OM?=AM.(OM-OM’). 
That istosay, (A4M-A0)?=AM.(4M+AM’-2A0), 
or AO?=AM.AM’. 
And, since QM?=p,.AM, and Q’M”"=p,.AM’, 
it follows that OME OM AO cc sae τονε, sates asada sare meeeeet ΠΟΥ (α). 
Now QV?: QD2=QP?: (ee) 


_QO’M'\2 
=Qv?: (ae “ἘΣ ) +QM.Q'M’ 
=QV?: (PN?+ QM. Q'M') 
=p.PV:pqg.(AN+4A0), by (a). 
But : PV=TO=AN+A0O. 
Therefore OV {ODl =p. Pas 


ON CONOIDS AND SPHEROIDS. 113 


Proposition 4. 
The area of any ellipse is to that of the aualiary circle as 
the minor axis to the major. 


Let AA’ be the major and BB’ the minor axis of the 
ellipse, and let BB’ meet the auxiliary circle in ὦ, δ΄. 


Suppose O to be such a circle that 
(circle ADA’D’): O= CA: CB. 
Then shall O be equal to the area of the ellipse. 
For, if not, O must be either greater or less than the 
ellipse. 
I. If possible, let O be greater than the ellipse. 


We can then inscribe in the circle O an equilateral polygon 
of 4n sides such that its area is greater than that of the ellipse. 
[cf. On the Sphere and Cylinder, I. 6.] 


Let this be done, and inscribe in the auxiliary circle of the 
ellipse the polygon AefbghA’... similar to that inscribed in Ὁ. 
Let the perpendiculars eM, fN,... on AA’ meet the ellipse in 
E, F,... respectively. Join AH, EF, FB.,.... 


Suppose that P’ denotes the area of the polygon inscribed 
in the auxiliary circle, and P that of the polygon inscribed in 
the ellipse. 


H. A. 8 


114 ARCHIMEDES 
Then, since all the lines eM, fN,... are cut in the same 
proportions at Δ’, F’,..., 
1.6. eM: EM=fN:FN=..,.=0C: BC, 
the pairs of triangles, as eA MW, HAM, and the pairs of trapeziums, 


as eMNf, EMNF, are all in the same ratio to one another 
as bC to BC, or as CA to CB. 


Therefore, by addition, 
Pe P= CA: CB. 
Now P’ : (polygon inscribed in Ὁ) 
= (circle AbA’b’): O 
= CA : CB, by hypothesis. 
Therefore P is equal to the polygon inscribed in Ὁ. 


But this is impossible, because the latter polygon is by 
hypothesis greater than the ellipse, and a fortiort greater 
than P. 


Hence O is not greater than the ellipse. 


II. If possible, let O be less than the ellipse. 


In this case we inscribe in the ellipse a polygon P with 4n 
equal sides such that P > 0. 


Let the perpendiculars from the angular points on the 
axis AA’ be produced to meet the auxiliary circle, and let the 
corresponding polygon (P’) in the circle be formed. 


Inscribe in O a polygon similar to P’, 

Then Py: P=CA»CB 
=(circle AbA’b’): O, by hypothesis, 
= P’: (polygon inscribed in 0). 


Therefore the polygon inscribed in O is equal to the 
polygon P; which is impossible, because P > 0. 


Hence O, being neither greater nor less than the ellipse, is 
equal to it; and the required result follows, 


ON CONOIDS AND SPHEROIDS. 11 


Proposition 5. 


If AA’, BB’ be the major and minor aais of an ellipse 
respectively, and if d be the diameter of any circle, then 
(area of ellipse) : (area of circle)= AA’. BB’: d?. 
For 
(area of ellipse) : (area of auxiliary circle) = BB’: AA’ [Prop. 4] 
ΞΡ. yal al 
And 
(area of aux. circle) : (area of circle with diam. d)= AA”: α΄. 


Therefore the required result follows ea aequali. 


Proposition 6. 


The areas of ellipses are as the rectangles under their axes. 
This follows at once from Props. 4, 5. 


Cor. The areas of similar ellipses are as the squares of 
corresponding axes. 


Proposition 7. 


Given an ellipse with centre C, and a line CO drawn per- 
pendicular to its plane, it ts possible to find a circular cone 
with vertex O and such that the given ellipse is a section of τί 
[or, in other words, to find the circular sections of the cone with 
vertex O passing through the circumference of the ellipse]. 


Conceive an ellipse with BB’ as its minor axis and lying in 
a plane perpendicular to that of the paper. Let CO be drawn 
perpendicular to the plane of the ellipse, and let O be the 
vertex of the required cone. Produce OB, OC, OB’, and in the 
same plane with them draw BHD meeting OC, OB’ produced 
in &, D respectively and in such a direction that 
Bie De BOF CA = CO", 
where CA is half the major axis of the ellipse. 


116 ARCHIMEDES 


“ And this is possible, since 
BE ED ~#O0;>BC.CB : ΟΟΣ 


[Both the construction and this proposition are assumed as 
known. | 


Now conceive a circle with BD as diameter lying in a plane 
at right angles to that of the paper, and describe a cone with 
this circle for its base and with vertex O. 


We have therefore to prove that the given ellipse is a 
section of the cone, or, if P be any point on the ellipse, that P 
lies on the surface of the cone. 


Draw PN perpendicular to BB’. Join ON and produce it 
to meet BD in M, and let MQ be drawn in the plane of the 
circle on BD as diameter perpendicular to BD and meeting the 
circle in Q. Also let FG, HK be drawn through 1, M respec- 
tively parallel to BB’. 


We have then 
QM?: HM.MkK=BM.MD: HM.MK 
=BE.ED: ΡῈ. Κα 
=(BE.ED : E0*).(HO’: FE. EG) 
= (CA* : CO*).(CO* : BC ΘΒ. 
= CA*: CB 
= PN*: BN.NB’. 


ON CONOIDS AND SPHEROIDS. Fey 


Therefore QM?: PN? =HM.MK:BN.NB’ 
=OM- SON: 
whence, since PN, QM are parallel, OPQ is a straight line. 


But Q is on the circumference of the circle on BD as 
diameter; therefore OQ is a generator of the cone, and hence 
P lies on the cone. 


Thus the cone passes through all points on the ellipse. 


Proposition 8. 


Given an ellipse, a plane through one of its axes AA’ and 
perpendicular to the plane of the ellipse, and a line CO drawn 
From C, the centre, in the given plane through AA’ but not 
perpendicular to AA’, it is possible to find a cone with vertex O 
such that the given ellipse is a section of i [or, in other words, 
to find the circular sections of the cone with vertex O whose 
surface passes through the circumference of the ellipse}. 


By hypothesis, 0A, OA’ are unequal. Produce 0.A’ to D so 
that OA=OD. Join AD, and draw FG through C parallel to it. 


The given ellipse is to be supposed to lie in a plane per- 
pendicular to the plane of the paper. Let BB’ be the other 
axis of the ellipse. 


Conceive a plane through AD perpendicular to the plane 
of the paper, and in it describe either (a), if CB’=FC.CG,a 
circle with diameter AD, or (b), if not, an ellipse on AD as 
axis such that, if d be the other axis, 


d?: AD? =CB’ : FC. CG. 


Take a cone with vertex 0 whose surface passes through 
the circle or ellipse just drawn. .This is possible even when the 
curve is an ellipse, because the line from O to the middle point 
of AD is perpendicular to the plane of the ellipse, and the 
construction is effected by means of Prop. 7. 


Let P be any point on the given ellipse, and we have only 
to prove that P lies on the surface of the cone so described, 


118 ARCHIMEDES 


Draw PN perpendicular to AA’. Join ON, and produce it 
to meet AD in M. Through M draw HK parallel to A’A. 


K 


Lastly, draw MQ perpendicular to the plane of the paper 
(and therefore perpendicular to both HK and AD) meeting the 
ellipse or circle about AD (and therefore the surface of the cone) 


in Q. 
Then 
QM’: HM.MK =(QM’: DM.MA).(DM.MA : HM.MK) 
=i - AD*). (C.CG ACAD 
=(CB* 2 FC.CG). fC. €G A C.Gay 
= CB": CA" 
= ΡΝ ἈΝ NA. 
Therefore, alternately, 
OMe: ἘΝ Ξ FM. Mike AGN a 
ΞΞ 0M OMe 
Thus, since PN, QM are parallel, OPQ is a straight line; 
and, @ being on the surface of the cone, it follows that P is also 
on the surface of the cone. 


Similarly all points on the ellipse are also on the cone, and 
the ellipse is therefore a section of the cone. 


ON CONOIDS AND SPHEROIDS. 119 


Proposition 9. 


Given an ellipse, a plane through one of its axes and perpen- 
dicular to that of the ellipse, and a straight line CO drawn from 
the centre C of the ellipse in the given plane through the axis but 
not perpendicular to that axis, it is possible to find a cylinder 
with axis OC such that the ellipse is a section of it [or, in other 
words, to find the circular sections of the cylinder with aais OC 
whose surface passes through the circumference of the given 
ellipse]. 

Let AA’ be an axis of the ellipse, and suppose the plane 
of the ellipse to be perpendicular to that of the paper, so that 
OC lies in the plane of the paper. 


see 


c NA 
Draw AD, A’E parallel to CO, and let DE be the line 
through O perpendicular to both AD and A’E. 


We have now three different cases according as the other 
axis BB’ of the ellipse is (1) equal to, (2) greater than, or 
(3) less than, DE. 

(1) Suppose BB’ = DE. 

Draw a plane through D# at right angles to OC, and in 
this plane describe a circle on DE as diameter. Through this 
circle describe a cylinder with axis OC. 

This cylinder shall be the cylinder required, or its surface 
shall pass through every point P of the ellipse. 

For, if P be any point on the ellipse, draw PN perpendicular 
to AA’; through NV draw NM parallel to CO meeting DE in 
M, and through M, in the plane of the circle on DE as diameter, 
draw MQ perpendicular to DE, meeting the circle in Q. 


120 ARCHIMEDES 


Then, since DE = BB’, 
PN AN Ne — DO" AC CAs 
And DM IME ANN Aa = D0": ACs 
since AD, NM, CO, A’E are parallel. 
Therefore PN*=DM.ME 
= QM’, 


by the property of the circle. 


Hence, since PN, QM are equal as well as parallel, PQ is 
parallel to MN and therefore to CO. It follows that PQ is a 
generator of the cylinder, whose surface accordingly passes 


through P. 

(2) If BB’> DE, we take E£’ on A’E such that DE’ = BB’ 
and describe a circle on DE’ as diameter in a plane perpen- 
dicular to that of the paper; and the rest of the construction 
and proof is exactly similar to those given for case (1). 

(3) Suppose BB’ < DE. 

Take a point K on CO produced such that 

DO* — CB’ = OR”. 

From K draw KR perpendicular to the plane of the paper 
and equal to CB. 

Thus OR? = OK? + CB’ = OD". 


A Cc N A’ 


In the plane containing DE, OR describe a circle on DE as 
diameter. Through this circle (which must pass through R) 
draw a cylinder with axis OC. 


ON CONOIDS AND SPHEROIDS. 191] 


We have then to prove that, if P be any point on the given 
ellipse, P lies on the cylinder so described. 

Draw PN perpendicular to AA’, and through V draw VM 
parallel to CO meeting DE in M. In the plane of the circle on 
DE as diameter draw MQ perpendicular to DE and meeting 
the circle in Q. 

Lastly, draw QH perpendicular to NM produced. QH will 
then be perpendicular to the plane containing AC, DE, 1.6. the 
plane of the paper. 


Now ΘΗ: QM’=KR?’ : OR’, by similar triangles. 
nnd OM AN NA’ = DM. ME: AN: NA’ 
=O. 6 Aq, 


Hence, ex aequali, since OR = OD, 
OH AN GNA = Ki: CAs 
= OB’ : CA’ 
= PN? AN NAS 
Thus Q7=PN. And ΟΠ, PN are also parallel. Accordingly 


PQ is parallel to MN, and therefore to CO, so that PQ is a 
generator, and the cylinder passes through P. 


Proposition 10. 


It was proved by the earlier geometers that any two cones 
have to one another the ratio compounded of the ratios of their 
bases and of their heights*. The same method of proof will 
show that any segments of cones have to one another the ratio 
compounded of the ratios of their bases and of their heights. 

The proposition that any ‘frustum’ of a cylinder is triple 
of the conical segment which has the same base as the frustum 
and equal height is also proved in the same manner as the 
proposition that the cylinder is triple of the cone which has 
the sume base as the cylinder and equal heightt. 


* This follows from Euel. x11. 11 and 14 taken together. Cf. On the Sphere 
and Cylinder τ, Lemma 1. 

+ This proposition was proved by Eudoxus, as stated in the preface to On 
the Sphere and Cylinder τ. Cf. Eucel. xr. 10. 


1. ARCHIMEDES 


Proposition 11. 


(1) Ifa paraboloid of revolution be cut by a plane through, 
or parallel to, the axis, the section will be a parabola equal to the 
original parabola which by its revolution generates the paraboloid. 
And the axis of the section will be the intersection between the 
cutting plane and the plane through the axis of the paraboloid 
at right angles to the cutting plane. 


If the paraboloid be cut by a plane at right angles to its 
axis, the section will be a circle whose centre is on the axis. 


(2) Ifa hyperboloid of revolution be cut by a plane through 
the axis, parallel to the axis, or through the centre, the section 
will be a hyperbola, (a) if the section be through the axis, equal, 
(0) af parallel to the axis, similar, (c) if through the centre, 
not similar, to the original hyperbola which by its revolution 
generates the hyperboloid. And the aais of the section will be 
the intersection of the cutting plane and the plane through the 
axis of the hyperboloid at right angles to the cutting plane. 


Any section of the hyperboloid by a plane at right angles to 
the axis will be a circle whose centre is on the axis. 


(3) Lf any of the spheroidal figures be cut by a plane through 
the axis or parallel to the axis, the section will be an ellipse, 
(a) of the section be through the axis, equal, (Ὁ) if parallel to the 
axis, similar, to the ellipse which by its revolution generates the 
jigure. And the axis of the section will be the intersection of the 
cutting plane and the plane through the aais of the spheroid 
at right angles to the cutting plane. 


If the section be by a plane at right angles to the axis of the 
spherovd, τέ will be a circle whose centre is on the axis. 


(4) If any of the said figures be cut by a plane through the 
axis, and if a perpendicular be drawn to the plane of section 
From any point on the surface of the figure but not on the section, 
that perpendicular will fall within the section. 

“ And the proofs of all these propositions are evident.” * 


* Cf. the Introduction, chapter 111. § 4. 


ON CONOIDS AND SPHEROIDS. ΡΞ 


Proposition 12. 


Tf a paraboloid of revolution be cut by a plane neither parallel 
nor perpendicular to the axis, and if the plane through the axis 
perpendicular to the cutting plane intersect it in a straight line 
of which the portion intercepted within the paraboloid is RR’, 
the section of the paraboloid will be an ellipse whose major axis 
is RR’ and whose minor axis is equal to the perpendicular 
distance between the lines through R, Ε΄ parallel to the axis 
of the paraboloid. 


Suppose the cutting plane to be perpendicular to the plane 
of the paper, and let the latter be the plane through the axis 
ANF of the paraboloid which intersects the cutting plane at 
right angles in RR’. Let RH be parallel to the axis of the 
paraboloid, and &’H perpendicular to RH. 

Let Q be any point on the section made by the cutting 
plane, and from @ draw QM perpendicular to RR’. QM will 
therefore be perpendicular to the plane of the paper. 

Through M draw DMFE perpendicular to the axis ANF 
meeting the parabolic section made by the plane of the paper 
in D, E. Then QM is perpendicular to D#, and, if a plane be 
drawn through DE, QM, it will be perpendicular to the axis 
and will cut the paraboloid in a circular section. 


Since Q is on this circle, 
QM’? = DM. ME. 
Again, if PT be that tangent to the parabolic section in the 


124 ARCHIMEDES 


plane of the paper which is parallel to RA’, and if the tangent 
at A meet PT in O, then, from the property of the parabola, 
DM .ME: RM.ME =A0*: OP? [Prop. 3 (1)] 
= (40"- OT’, since AN= AT. 
Therefore QM’: RM.MR'= AO’: OT” 
=H”. RR 
by similar triangles. 
Hence ᾧ lies on an ellipse whose major axis is RR’ and 
whose minor axis is equal to {΄ ἢ. 


Propositions 13, 14. 


If a hyperboloid of revolution be cut by a plane meeting all 
the generators of the enveloping cone, or if an ‘ oblong’ spheroid 
be cut by a plane not perpendicular to the axis*, and if a plane 
through the axis intersect the cutting plane at right angles in a 
straight line on which the hyperboloid or spheroid intercepts 
a length RR’, then the section by the cutting plane will be an 
ellipse whose major axis is RR’. 


Suppose the cutting plane to be at right angles to the 
plane of the paper, and suppose the latter plane to be that 


* Archimedes begins Prop, 14 for the spheroid with the remark that, when the 
cutting plane passes through or is parallel to the axis, the case is clear (δῆλον). 
Cf. Prop. 11 (8). 


ON CONOIDS AND SPHEROIDS. 125 


through the axis ANF which intersects the cutting plane 
at right angles in RR’. The section of the hyperboloid or 
spheroid by the plane of the paper is thus a hyperbola or ellipse 
having AF for its transverse or major axis. 

Take any point on the section made by the cutting plane, 
as Ὁ, and draw QM perpendicuiar to RR’. QM will then 
be perpendicular to the plane of the paper. 

Through M draw DFE at right angles to the axis ANF 
meeting the hyperbola or ellipse in D, #; and through QM, 
DE \et a plane be described. This plane will accordingly be 
perpendicular to the axis and will cut the hyperboloid or 
spheroid in a circular section. 


Thus QM?=DM.ME. 


Let PT be that tangent to the hyperbola or ellipse which 
is parallel to RR’, and let the tangent at A meet PT in Ὁ. 


Then, by the property of the hyperbola or ellipse, 
DV ME ie Mi: = OA OP" 
or QM’: RM. MR’ = OA’: OP’, 


Now (1) in the hyperbola OA < OP, because 47 < AN*, and 
accordingly O7'< OP, while OA < OT, 


(2) in the ellipse, if A.’ be the diameter parallel to RR’, 
and BB’ the minor axis, 


BORO BIC Clg = OA O PY 
and BC.CB’< KC.CK’, so that OA < OP. 
Hence in both cases the locus of @ is an ellipse whose major 
axis is RR’. 
Cor. 1. If the spheroid be a ‘flat’ spheroid, the section will 


be an ellipse, and everything will proceed as before except that 
RR’ will in this case be the minor axis. 


Cor. 2. In all conoids or spheroids parallel sections will be 
similar, since the ratio OA?: ΟΡ" is the same for all the 
parallel sections. 


* With reference to this assumption cf. the Introduction, chapter ur. § 3. 


126 ARCHIMEDES 


Proposition 15. 


(1) Lf from any point on the surface of a conoid a line be 
drawn, in the case of the paraboloid, parallel to the axis, and, in 
the case of the hyperboloid, parallel to any line passing through 
the vertex of the enveloping cone, the part of the straight line 
which is in the same direction as the convexity of the surface will 
fall without it, and the part which is in the other direction 
within τέ. 


For, if a plane be drawn, in the case of the paraboloid, 
through the axis and the point, and, in the case of the hyperbo- 
loid, through the given point and through the given straight 
line drawn through the vertex of the enveloping cone, the 
section by the plane will be (a) in the paraboloid a parabola 
whose axis is the axis of the paraboloid, (b) in the hyperboloid 
a hyperbola in which the given line through the vertex of the 
enveloping cone is a diameter *. [ Prop. 11] 


Hence the property follows from the plane properties of the 
conics. 

(2) If a plane touch a conoid without cutting rt, τέ will 
touch it at one point only, and the plane drawn through the 
point of contact and the aais of the conoid will be at right 
angles to the plane which touches τί. 


For, if possible, let the plane touch at two points. Draw 
through each point a parallel to the axis. The plane passing 
through both parallels will therefore either pass through, or be 
parallel to, the axis. Hence the section of the conoid made by 
this plane will be a conic [Prop. 11 (1), (2)], the two points 
will lie on this conic, and the line joining them will lie within 
the conic and therefore within the conoid. But this lne 
will be in the tangent plane, since the two points are in it. 
Therefore some portion of the tangent plane will be within 
the conoid; which is impossible, since the plane does not 
cut it. 


* There seems to be some error in the text here, which says that ‘the 
diameter” (i.e. axis) of the hyperbola is ‘‘ the straight line drawn in the conoid 
from the vertex of the cone.’’ But this straight line is not, in general, the 
axis of the section. 


ON CONOIDS AND SPHEROIDS. 127 


Therefore the tangent plane touches in one point only. 


That the plane through the point of contact and the axis is 
perpendicular to the tangent plane is evident in the particular 
case where the point of contact is the vertex of the conoid. 
For, if two planes through the axis cut it in two conics, the 
tangents at the vertex in both conics will be perpendicular 
to the axis of the conoid. And all such tangents will be in the 
tangent plane, which must therefore be perpendicular to the 
axis and to any plane through the axis. 


If the point of contact P is not the vertex, draw the plane 
passing through the axis AWN and the point P. 
It will cut the conoid in a conic whose axis is 
AN and the tangent plane in a line DPE 
touching the conic at P. Draw PNP’ perpen- 
dicular to the axis, and draw a plane through it 
also perpendicular to the axis. This plane will 
make a circular section and meet the tangent 
plane in a tangent to the circle, which will 
therefore be at nght angles to PN. Hence the 
tangent to the circle will be at mght angles to the plane 
containing PN, AN; and it follows that this last plane is 
perpendicular to the tangent plane. 


Ε 


Proposition 16. 


(1) 1 a plane touch any of the spheroidal figures without 
cutting it, it will touch at one point only, and the plane through 
the point of contact and the axis will be at right angles to the 
tangent plane. 


This is proved by the same method as the last proposition. 


(2) If any conoid or spheroid be cut by a plane through the 
axis, and if through any tangent to the resulting conic a plane be 
erected at right angles to the plane of section, the plane so erected 
will touch the conoid or spheroid in the same point as that vn 
which the line touches the conic. 


For it cannot meet the surface at any other point. If it 
did, the perpendicular from the second point on the cutting 


128 ARCHIMEDES 


plane would be perpendicular also to the tangent to the conic 
and would therefore fall outside the surface. But it must fall 
within it. [Prop. 11 (4)] 


(3) If two parallel planes touch any of the spheroidal 
figures, the line joining the points of contact will pass through 
the centre of the spherovd. 


If the planes are at right angles to the axis, the proposition 
is obvious. If not, the plane through the axis and one point of 
contact is at right angles to the tangent plane at that point. 
It is therefore at right angles to the parallel tangent plane, and 
therefore passes through the second point of contact. Hence 
both points of contact lie on one plane through the axis, and 
the proposition is reduced to a plane one. 


Proposition 17. 


If two parallel planes touch any of the spheroidal figures, 
and another plane be drawn parallel to the tangent planes and 
passing through the centre, the line drawn through any point of 
the circumference of the resulting section parallel to the chord 
of contact of the tangent planes will fall outside the spheroid. 


This is proved at once by reduction to a plane proposition. 


Archimedes adds that it is evident that, if the plane 
parallel to the tangent planes does not pass through the 
centre, a straight line drawn in the manner described will 
fall without the spheroid in the direction of the smaller 
segment but within it in the other direction. 


Proposition 18. 


Any spheroidal figure which is cut by a plane through the 
centre is divided, both as regards its surface and its volume, into 
two equal parts by that plane. 


To prove this, Archimedes takes another equal and similar 
spheroid, divides it similarly by a plane through the centre, and 
then uses the method of application. 


ON CONOIDS AND SPHEROIDS. 129 


Propositions 19, 20. 


Given a segment cut off by a plane from a paraboloid or 
hyperboloid of revolution, or a segment of a spheroid less than 
half the spheroid also cut off by a plane, τέ 1s possible to inscribe 
in the segment one solid figure and to circumscribe about it 
another solid figure, each made up of cylinders or ‘frusta’ of 
cylinders of equal height, and such that the circumscribed figure 
exceeds the inscribed figure by a volume less than that of any 
gwen solid. 

Let the plane base of the segment be perpendicular to the 
plane of the paper, and let the plane of the paper be the plane 
through the axis of the conoid or spheroid which cuts the base 
of the segment at right angles in BC. The section in the plane 
of the paper is then a conic BAC. [Prop. 11] 

Let HAF be that tangent to the conic which is parallel to 
BC, and let A be the point of contact. Through HAF draw 
a plane parallel to the plane through BC bounding the 
segment. The plane so drawn will then touch the conoid 
or spheroid at A. [Prop. 16] 

(1) If the base of the segment is at right angles to the 
axis of the conoid or spheroid, A will be the vertex of the 
conoid or spheroid, and its axis AD will bisect BC at right 
angles. 

(2) Ifthe base of the segment is not at right angles to the 
axis of the conoid or spheroid, we draw AD 


(a) in the paraboloid, parallel to the axis, 

(6) in the hyperboloid, through the centre (or the vertex of 
the enveloping cone), 

(c) in the spheroid, through the centre, 
and in all the cases it will follow that AD bisects BC in D. 


Then A will be the vertex of the segment, and AD will be 
itS axis. 


Further, the base of the segment will be a circle or an 
ellipse with BC as diameter or as an axis respectively, and 
with centre D. We can therefore describe through this circle 


H. A. 9 


130 ARCHIMEDES 


or ellipse a cylinder or a ‘frustum’ of a cylinder whose axis is 


AD. [Prop. 9] 


Dividing this cylinder or frustum continually into equal 
parts by planes parallel to the base, we shall at length arrive 
at a cylinder or frustum less in volume than any given solid. 

Let this cylinder or frustum be that whose axis is OD, and 
let AD be divided into parts equal to OD, at L, M,..... Through 
L, M.,... draw lines parallel to BC meeting the conic in P, Q...., 
and through these lines draw planes parallel to the base of the 
segment. These will cut the conoid or spheroid in circles or 
similar ellipses. On each of these circles or ellipses describe 
two cylinders or frusta of cylinders each with axis equal to OD, 
one of them lying in the direction of A and the other in the 
direction of D, as shown in the figure. 

Then the cylinders or frusta of cylinders drawn in the 
direction of A make up a circumscribed figure, and those in 
the direction of D an inscribed figure, in relation to the 
segment. 

Also the cylinder or frustum PG in the circumscribed figure 
is equal to the cylinder or frustum P# in the inscribed figure, 
QJ in the circumscribed figure is equal to QK in the inscribed 
figure, and so on. 

Therefore, by addition, 

(circumscribed fig.) = (inscr, fig.) 
+ (cylinder or frustum whose axis is OD), 

But the cylinder or frustum whose axis is OD is less than 
the given solid figure ; whence the proposition follows. 


“Having set out these preliminary propositions, let us 
proceed to demonstrate the theorems propounded with reference 
to the figures.” 


ON CONOIDS AND SPHEROIDS. 131 


Propositions 21, 22. 


Any segment of a paraboloid of revolution is half as large 
again as the cone or segment of a cone which has the same base 
and the same amis. 


Let the base of the segment be perpendicular to the plane of 
the paper, and let the plane of the paper be the plane through 
the axis of the paraboloid which cuts the base of the segment 
at right angles in BC and makes the parabolic section BAC. 


Let HF be that tangent to the parabola which is parallel to 
BC, and let A be the point of contact. 


Then (1), if the plane of the base of the segment is 
perpendicular to the axis of the paraboloid, that axis is the 
line AD bisecting BC at right angles in D. 


(2) If the plane of the base is not perpendicular to the 
axis of the paraboloid, draw AD parallel to the axis of the 
paraboloid. AD will then bisect BC, but not at right angles. 

Draw through #F a plane parallel to the base of the seg- 
ment. This will touch the paraboloid at A, and A will be 
the vertex of the segment, AD its axis. 

The base of the segment will be a circle with diameter BC 
or an ellipse with BC as major axis. 

Accordingly a cylinder or a frustum of a cylinder can be 
found passing through the circle or ellipse and having AD for 
its axis | Prop. 9]; and likewise a cone or a segment of a cone 
can be drawn passing through the circle or ellipse and having 
A for vertex and AD for axis. [ Prop. 8] 

Suppose X to be a cone equal to 3 (cone or segment of 
cone ABC). The cone X is therefore equal to half the cylinder 
or frustum of a cylinder EC. [ΟἹ Prop. 10] 

We shall prove that the volume of the segment of the 
paraboloid is equal to X. 

If not, the segment must be either greater or less than X. 

I. If possible, let the segment be greater than YX. 

We can then inscribe and circumscribe, as in the last 

9—2 


132 ARCHIMEDES 


proposition, figures made up of cylinders or frusta of cylinders 
with equal height and such that 
(circumscribed fig.) — (inscribed fig.) < (segment) — X. 

Let the greatest of the cylinders or frusta forming the 
circumscribed figure be that whose base is the circle or ellipse 
about BC and whose axis is OD, and let the smallest of them be 
that whose base is the circle or ellipse about PP’ and whose 
axis is AL. 

Let the greatest of the cylinders forming the inscribed 
figure be that whose base is the circle or ellipse about RR’ and 
whose axis is OD, and let the smallest be that whose base is 
the circle or ellipse about PP’ and whose axis is LM. 


Produce all the plane bases of the cylinders or frusta to 
meet the surface of the complete cylinder or frustum £C, 
Now, since 
(circumscribed fig.) — (inscr, fig.) < (segment) — ἃ, 
it follows that (inscribed figure)'> X 22.6.5. .sesueeee (a). 
Next, comparing successively the cylinders or frusta with 
heights equal to OD and respectively forming parts of the 
complete cylinder or frustum #C and of the inscribed figure, 
we have 
(first cylinder or frustum in /(C) : (first in inscr. fig.) 
= BD*: 0: 
Pe ean 
= BD: TO, where AB meets OR in 7. 
And (second cylinder or frustum in HC) : (second in inscr. fig.) 
= HO: SN, in like manner, 


and so on. 


ON CONOIDS AND SPHEROIDS. roo 


Hence [Prop. 1] (cylinder or frustum ZC) : (inscribed figure) 
= (BD+ HO+...):(T0+S8SN+...), 
where BD, HO,... are all equal, and BD, TO, SN,... diminish in 
arithmetical progression. 
But [Lemma preceding Prop. 1] 
BD+H0O+...>2(TO+SN+...). 
Therefore (cylinder or frustum HC) > 2 (inscribed fig.), 
or X > (inscribed fig.) ; 
which is impossible, by (a) above. 
II. If possible, let the segment be less than YX. 
In this case we inscribe and circumscribe figures as before, 
but such that 
(circumscer. fig.) — (inser. fig.) « X — (segment), 
whence it follows that 
(circumscribed figure) < X ............56. (8). 
And, comparing the cylinders or frusta making up the 
complete cylinder or frustum CE and the circumscribed figure 
respectively, we have | 
(first cylinder or frustum in CZ) : (first in circumscr. fig.) 
=) (2 8» 
— BD Bil). 
(second in CZ) : (second in cireumscr. fig.) 
=O" 7 RO" 
ΞΞ AO 
ΞΞΠ ΠΣ 
and so on. 
Hence [Prop. 1] 
(cylinder or frustum CZ) : (circumscribed fig.) 
=(BD+ HO +...):(BD+TO+...), 
«Ὁ 01. [Lemma preceding Prop. 1] 
and it follows that 
X < (circumscribed fig.) ; 
which is impossible, by (8). 
Thus the segment, being neither greater nor less than Δ΄, is 
equal to it, and therefore to 3 (cone or segment of cone ABC). 


134 ARCHIMEDES 


Proposition 23. 


If from a paraboloid of revolution two segments be cut off, 
one by a plane perpendicular to the amis, the other by a plane not 
perpendicular to the axis, and if the axes of the segments are 
equal, the segments will be equal in volume. 


Let the two planes be supposed perpendicular to the plane 
of the paper, and let the latter plane be the plane through the 
axis of the paraboloid cutting the other two planes at right 
angles in BB’, QQ’ respectively and the paraboloid itself in the 
parabola QPQ’B'. 

Let AN, PV be the equal axes of the segments, and A, P 
their respective vertices. 


Draw QL parallel to AN or PV and Q’L perpendicular 
to QL. 


Now, since the segments of the parabolic section cut off by 
BB’, QQ’ have equal axes, the triangles ABB’, PQQ’ are equal 
[Prop. 3]. Also, if QD be perpendicular to PV, QD= BN (as 
in the same Prop. 3). 

Conceive two cones drawn with the same bases as the 
segments and with A, P as vertices respectively. The height 
of the cone PQQ’ is then PK, where PK is perpendicular to 
QQ". 


ON CONOIDS AND SPHEROIDS. 135 


Now the cones are in the ratio compounded of the ratios of 
their bases and of their heights, 1.6. the ratio compounded of 
(1) the ratio of the circle about BB’ to the ellipse about QQ’, 
and (2) the ratio of AN to PK. 


That is to say, we have, by means of Props. 5, 12, 
(cone ABB’): (cone PQQ) =(BB" : QQ’. Q'L).(AN : PR). 
And BB’ =26N—20D— 0 L, while QQ’ =2QV. 
Therefore 
(cone ABB’) : (cone PQQ’) =(QD: QV).(AN : PRK) 

Ξε PV) «AN = PR) 
— AN es 

Since AN = PY, the ratio of the cones is a ratio of equality ; 


and it follows that the segments, being each half as large again 
as the respective cones [Prop. 22], are equal, 


Proposition 24. 


If from a paraboloid of revolution two segments be cut off by 
planes drawn in any manner, the segments will be to one another 
as the squares on their ames. 


For let the paraboloid be cut by a plane through the axis 
in the parabolic section P’PApp’, and let the axis of the 
parabola and paraboloid be ANN’. 

Measure along ANN’ the lengths AN, AN’ equal to the 
respective axes of the given segments, 
and through V, V’ draw planes perpen- ek 
dicular to the axis, making circular 
sections on Pp, P’p’ as diameters re- 
spectively. With these circles as bases 
and with the common vertex A let two 
cones be described. 


Now the segments of the paraboloid 
whose bases are the circles about Pp, 
P’p’ are equal to the given segments 
respectively, since their respective axes 
are equal [Prop. 23]; and, since the 
segments APp, AP’p’ are half as large 


136 ARCHIMEDES 


again as the cones APp, AP’p’ respectively, we have only 
to show that the cones are in the ratio of AN’ to AN”. 


But 

(cone APp) : (cone AP’p’)=(PN?: P’N”).(AN : AN) 
=(AN: AN’).(AN: AN’) 
ΞΞ 2 AN ee 


thus the proposition is proved. 


Propositions 25, 26. 


In any hyperboloid of revolution, if A be the vertex and AD 
the axis of any segment cut off by a plane, and if CA be the 
semidiameter of the hyperboloid through A (CA being of course 
in the same straight line with AD), then 


(segment) : (cone with same base and axis) 
=(AD+3CA): (AD + 2CA). 


Let the plane cutting off the segment be perpendicular to 
the plane of the paper, and let the latter plane be the plane 
through the axis of the hyperboloid which intersects the cutting 
plane at right angles in BB’, and makes the hyperbolic 
segment BAB’. Let C be the centre of the hyperboloid (or 
the vertex of the enveloping cone). 


Let HF be that tangent to the hyperbolic section which is 
parallel to BB’. Let HF touch at A, and join CA. Then CA 
produced will bisect BB’ at D, CA will be a semi-diameter of 
the hyperboloid, A will be the vertex of the segment, and AD 
its axis. Produce AC to A’ and H, so that AC=CA’=A’H. 


Through #F draw a plane parallel to the base of the seg- 
ment. This plane will touch the hyperboloid at A. 


Then (1), if the base of the segment is at right angles to the 
axis of the hyperboloid, A will be the vertex, and AD the axis, 
of the hyperboloid as well as of the segment, and the base of the 
segment will be a circle on BB’ as diameter. 


ON CONOIDS AND SPHEROIDS. 1590 


(2) If the base of the segment is not perpendicular to the 
axis of the hyperboloid, the base will be an ellipse on BB’ as 
major axis. [Prop. 13] 


H 


Then we can draw a cylinder or a frustum of a cylinder 
EBB’F passing through the circle or ellipse about BB’ and 
having AD for its axis; also we can describe a cone or a 
segment of a cone through the circle or ellipse and having A 
for its vertex. 


We have to prove that 
(segment ABB’): (cone or segment of cone ABB’)= HD: A’D. 


138 ARCHIMEDES 


Let V be a cone such that 
V : (cone or segment of cone ABB’)= HD: A’D....... (a) 
and we have to prove that V is equal to the segment. 
Now 
(cylinder or frustum /B’) : (cone or segmt. of cone A BB’) =3 :1. 


Therefore, by means of (a), 


(cylinder or frustum HB’): V=A’'D: “= ΒΤ. (8). 


If the segment is not equal to V, it must either be greater 
or less. 


I. If possible, let the segment be greater than V. 


Inscribe and circumscribe to the segment figures made up 
of cylinders or frusta of cylinders, with axes along AD and all 
equal to one another, such that 


(circumscribed fig.) — (inser. fig.) « (segmt.) — V, 
whence Gmseribed figure) > V ...asi<c.neeanee (γ). 


Produce all the planes forming the bases of the cylinders or 
frusta of cylinders to meet the surface of the complete cylinder 
or frustum 4B’. 


Then, if VD be the axis of the greatest cylinder or frustum 
in the circumscribed figure, the complete cylinder will be 
divided into cylinders or frusta each equal to this greatest 
cylinder or frustum. 


Let there be a number of straight lines a equal to 4A’ and 
as many in number as the parts into which AD is divided by 
the bases of the cylinders or frusta. To each line a apply a 
rectangle which shall overlap it by a square, and let the greatest 
of the rectangles be equal to the rectangle AD. A’D and the 
least equal to the rectangle AL .A’L; also let the sides of the 
overlapping squares ὦ, p, 4,...« be in descending arithmetical 
progression. Thus ὦ, p, q,...l will be respectively equal to AD, 
AN, AM,...AL, and the rectangles (ab + 05), (ap + p’),...(al + 1) 
will be respectively equal to AD. A’D, AN. A’N,...AL. 411. 


ON CONOIDS AND SPHEROIDS. 139 


Suppose, further, that we have a series of spaces S each 
equal to the largest rectangle AD. Α΄ ἢ and as many in number 
as the diminishing rectangles. 


Comparing now the successive cylinders or frusta (1) in the 
complete cylinder or frustum #B’ and (2) in the inscribed 
figure, beginning from the base of the segment, we have 


(first cylinder or frustum in 187) : (first in inscr, figure) 
= bee 
=AD.A'D: AN. A’N, from the hyperbola, 
=S:(apt+p’). 


Again 
(second cylinder or frustum in 487) : (second in inser. fig.) 
ΞΘ EON 
ΞΞ eee Di AM AM 
Ξ- 5: (aq Ὁ 4), 


and so on. 


The last cylinder or frustum in the complete cylinder or 
frustum HB’ has no cylinder or frustum corresponding to it in 
the inscribed figure. 


Combining the proportions, we have [Prop. 1] 
(cylinder or frustum #6’): (inscribed figure) 
= (sum of all the spaces 8S): (ap + p*)+(aq+q°) +... 


>(a+b): (Ξ - 3) [Prop. 2] 


Sy ON. sincea=AA’, b= AD, 


>(#B’): V, by (8B) above. 
Hence (inscribed figure) < V. 


But this is impossible, because, by (vy) above, the inscribed 
figure is greater than V. 


140 ARCHIMEDES 


II. Next suppose, if possible, that the segment is less 
than V. 


In this case we circumscribe and inscribe figures such that 
(circumscribed fig.) — (inscribed fig.) « V— (segment), 
whence we derive 
V > (circumscribed figure) ...........+0+. (8). 


We now compare successive cylinders or frusta in the 
complete cylinder or frustum and in the circumscribed figure ; 
and we have 


(first cylinder or frustum in £B’) : (first in circumscribed fig.) 


==) ous) 
=S: (ab - 65), 
(second in #B’) : (second in circumscribed fig.) 
‘ =S: (ap +p’), 


and so on. 
Hence [Prop. 1] 
(cylinder or frustum /#B’) : (circumscribed fig.) 
=(sum of all spaces S) : (ab + 05) + (ap+p*)+... 


<(a+b): (: + 3) [Prop. 2] 


< AD): aD 
3 
<(EB’): V, by (8) above. 


Hence the circumscribed figure is greater than V; which is 
impossible, by (δ) above. 


Thus the segment is neither greater nor less than V, and is 
therefore equal to it. 


Therefore, by (a), 


(segment ABB’): (cone or segment of cone ABB’) 
=(AD+3CA): (AD+ 2CA). 


ON CONOIDS AND SPHEROIDS. 141 


Propositions 27, 28, 29, 30. 


(1) In any spheroid whose centre is C, of a plane meeting 
the axis cut off a segment not greater than half the spheroid and 
having A for its vertex and AD for its axis, and if A’D be the 
axis of the remaining segment of the spheroid, then 


(first segmt.) : (cone or segmt. of cone with same base and axis) 
=CA+A’D: A’D 
[=3CA —AD: 2CA — AD]. 


(2) As a particular case, if the plane passes through the 
centre, so that the segment is half the spheroid, half the spheroid 
us double of the cone or segment of a cone which has the same 
vertex and axis. 


Let the plane cutting off the segment be at right angles to 
the plane of the paper, and let the latter plane be the plane 
through the axis of the spheroid which intersects the cutting 
plane in BB’ and makes the elliptic section A BA’B’. 


Let EF, Μ΄ Κ΄ be the two tangents to the ellipse which are 
parallel to BB’, let them touch it in A, A’, and through the 
tangents draw planes parallel to the base of the segment. 
These planes will touch the spheroid at A, A’, which will 
be the vertices of the two segments into which it is divided. 
Also AA’ will pass through the centre C and bisect BB’ 
in D. 


Then (1) if the base of the segments be perpendicular to 
the axis of the spheroid, A, A’ will be the vertices of the 
spheroid as well as of the segments, AA’ will be the axis 
of the spheroid, and the base of the segments will be a circle on 
BB’ as diameter ; 


(2) if the base of the segments be not perpendicular to the 
axis of the spheroid, the base of the segments will be an 
ellipse of which BB’ is one axis, and AD, A’D will be the 
axes of the segments respectively. 


142 ARCHIMEDES 


We can now draw a cylinder or a frustum of a cylinder 
EBB’F through the circle or ellipse about BB’ and having AD 
for its axis; and we can also draw a cone or a segment of 
a cone passing through the circle or ellipse about BB’ and 
having A for its vertex. 


We have then to show that, if CA’ be produced to H so 
ἘΠῚ ΟΣ 5 = ACH, 
(segment A BB’) : (cone or segment of cone ABB’) = HD: A’D. 
Let V be such a cone that 
V : (cone or segment of cone ABB’)= HD: A’D ... (a); 
and we have to show that the segment ABB’ is equal to V. 


ON CONOIDS AND SPHEROIDS. 143 


But, since 


(cylinder or frustum #B’) : (cone or segment of cone ABB’) 
ΞΟ 
we have, by the aid of (a), 


(cylinder or frustum #B’): V=A’'D: 5. (β). 


Now, if the segment ABB’ is not equal to V, it must 
be either greater or less. 


I. Suppose, if possible, that the segment is greater 
than V. 


Let figures be inscribed and circumscribed to the segment 
consisting of cylinders or frusta of cylinders, with axes along 
AD and all equal to one another, such that 

(circumscribed fig.) — (inscribed fig.) < (segment) — V, 
whence it follows that 


{περι ρει fig) συ... (γ). 


Produce all the planes forming the bases of the cylinders or 
frusta to meet the surface of the complete cylinder or frustum 
EB’. Thus, if ND be the axis of the greatest cylinder or 
frustum of a cylinder in the circumscribed figure, the complete 
cylinder or frustum Δ Β΄ will be divided into cylinders or frusta 
of cylinders each equal to the greatest of those in the circum- 


scribed figure. 


Take straight lines da’ each equal to A’D and as many in 
number as the parts into which AD is divided by the bases of 
the cylinders or frusta, and measure da along da’ equal to AD. 
It follows that aa’ = 2CD. 


Apply to each of the lines a’d rectangles with height equal 
to ad, and draw the squares on each of the lines ad as in 
the figure. Let S denote the area of each complete rectangle. 


From the first rectangle take away a gnomon with breadth 
equal to AWN (i.e. with each end of a length equal to AN); 
take away from the second rectangle a gnomon with breadth 
equal to AM, and so on, the last rectangle having no gnomon 
taken from it. 


144 ARCHIMEDES 


Then 
the first gnomon = A’D. AD—ND.(A'D-— AN) 
=A’'D.AN+ND.AN 
=AN.A’N. 
Similarly, 
the second gnomon = AM. A’M, 
and so on. 

And the last gnomon (that in the last rectangle but one) is 
equal to AL. A’L. 

Also, after the gnomons are taken away from the successive 
rectangles, the remainders (which we will call R,, R,,... Rn, 
where ἢ is the number of rectangles and accordingly R, Ξε 5) 
are rectangles applied to straight lines each of length aa’ and 
“exceeding by squares” whose sides are respectively equal 
to DN, DM,... Day 

For brevity, let DN be denoted by ὦ, and aa’ or 2CD by ο, 
so that R, =ca+a’, R,=c.2a+ (22)’,... 

Then, comparing successively the cylinders or frusta of 
cylinders (1) in the complete cylinder or frustum HB’ and 
(2) in the inscribed figure, we have 

(first cylinder or frustum in £'B’) : (first in inscribed fig.) 

= Bp* : PN* 
=AD.A'D: AN.A'N 
= 8S: (first gnomon) ; 
(second cylinder or frustum in #B’): (second in inscribed fig.) 
= S : (second gnomon), 
and so on. 

The last of the cylinders or frusta in the cylinder or 
frustum #B' has none corresponding to it in the inscribed 
figure, and there is no corresponding gnomon. 

Combining the proportions, we have [by Prop. 1] 

(cylinder or frustum 4.8} : (inscribed fig.) 
= (sum of all spaces S) : (sum of gnomons). 


ON CONOIDS AND SPHEROIDS. 145 


Now the differences between S and the successive gnomons 
are ἦν, H,,... Rn, while 
R, = οὐ + 2’, 
R,=c. 2a + (22), 


ΠΤ ΨΚ ΨΚ ΨΞ.Ί.Ί. 


ten COe ss, 
where ὦ = nz= AD. 
Hence [Prop. 2] 
(sum of all spaces 8S): (R,+Rh,+...+ Rn) « (ὁ -- ὁ): (5 -- a 
It follows that 
(sum of all spaces S) : (sum of gnomons) >(c + ὁ): “ = - 
> AD: ἘΠ᾿ 
.) 
Thus (cylinder or frustum 4B’) : (inscribed fig.) 
τὰν 


> (cylinder or frustum 2B’): V, 

from (8) above. 

Therefore (inscribed fig.) < V; 
which is impossible, by (ry) above. 

Hence the segment ABB’ is not greater than V. 

II. If possible, let the segment ABB’ be less than JV. 

We then inscribe and circumscribe figures such that 

(circumscribed fig.) — (inscribed fig.)< V — (segment), 
whence V > (circumsenibed fig). ..:..3.<220e 00s (8). 

In this case we compare the cylinders or frusta in (#'B’) 


with those in the circumscribed figure. 


Thus 
(first cylinder or frustum in #'B’) : (first in circumscribed fig.) 


Sf) SIE 
(second in £B’) : (second in circumscribed fig.) 
=: (first gnomon), 
and so on. 
Η. A, 10 


146 ARCHIMEDES 


Lastly (last in #B’) : (last in circumscribed fig.) 


=: (last gnomon). 
Now 


{S +(all the gnomons)} = nS — (A, + R,+...+ Ue) 


b 
οι... τ (5+5): [Prop. 2] 
so that 


nS : {S - (411 the gnomons)} « (ὁ - δ): (5 Ξε =) : 


It follows that, if we combine the above proportions as in 
Prop. 1, we obtain 


(cylinder or frustum /B’) : (circumscribed fig.) 
_(¢ , 2b. 
<(c+0): τς, 


ee 


<(EB’): V, by (8) above. 
Hence the circumscribed figure is greater than V; which is 
impossible, by (6) above. 


Thus, since the segment ABS’ is neither greater nor less 
than V, it is equal to it; and the proposition is proved. 


_ (2) The particular case [Props. 27, 28] where the segment 
is half the spheroid differs from the above in that the distance 
CD or c/2 vanishes, and the rectangles cb + b’ are simply squares 


(05), so that the gnomons are simply the differences between δ᾽ 
and 2”, δ᾽ and (2)’, and so on. 


Instead therefore of Prop. 2 we use the Lemma to Prop. 2, 
Cor. 1, given above [On Spirals, Prop. 10], and instead of the 


ratio (ὁ + ὁ): ᾿ + = we obtain the ratio 3 : 2, whence 


(segment ABB’) : (cone or segment of cone ABB’) =2 : 1. 


[This result can also be obtained by simply substituting 
CA for AD in the ratio (364 — AD) : (2CA — AD).] 


ON CONOIDS AND SPHEROIDS. 147 


Propositions 31, 32. 


If a plane divide a spheroid into two unequal segments, and 
uf AN, A’N be the axes of the lesser and greater segments 
respectively, while C is the centre of the spheroid, then 
(greater segmt.) : (cone or segmt. of cone with same base and axis) 

ΞΟΑΈΑΝ: ΑΝ. 

Let the plane dividing the spheroid be that through PP’ 
perpendicular to the plane of the paper, and let the latter plane 
be that through the axis of the spheroid which intersects the 
cutting plane in PP’ and makes the elliptic section PAP’A’. 


Draw the tangents to the ellipse which are parallel to PP’; 
let them touch the ellipse at A, A’, and through the tangents 
draw planes parallel to the base of the segments. These planes 
will touch the spheroid at <A, A’, the line AA’ will pass 
through the centre C and bisect PP’ in N, while AN, A’N will 
be the axes of the segments. 

Then (1) if the cutting plane be perpendicular to the axis 
of the spheroid, AA’ will be that axis, and A, A’ will be the 
vertices of the spheroid as well as of the segments. Also the 
sections of the spheroid by the cutting plane and all planes 
parallel to it will be circles. 


(2) If the cutting plane be not perpendicular to the axis, 
10—2 


148 ARCHIMEDES 


the base of the segments will be an ellipse of which PP’ is an 
axis, and the sections of the spheroid by all planes parallel 
to the cutting plane will be similar ellipses. 


Draw a plane through C parallel to the base of the segments 
and meeting the plane of the paper in BB’. 


Construct three cones or segments of cones, two having A 
for their common vertex and the plane sections through PP’, 
BB’ for their respective bases, and a third having the plane 
section through PP’ for its base and A’ for its vertex. 


Produce CA to H and CA’ to H’ so that 
AH =A'H’— CA. 
We have then to prove that 
(segment A’PP’) : (cone or segment of cone A’PP’) 
=CA+ AN - Am 
= Vd Aue 


Now half the spheroid is double of the cone or segment of a 
cone ABB’ [Props. 27, 28]. Therefore 


(the spheroid) = 4 (cone or segment of cone ABB’). 
But 
(cone or segmt. of cone ABB’): (cone or segmt. of cone APP’) 
=(CA AN) BC PING) 
=(CA:AN).(CA CA): AN. AW) ΩΣ 
If we measure AK along AA’ so that 
AK: AC=AC: AW, 
we have AK .A'N:AC.A’N=CA: AN, 
and the compound ratio in (a) becomes 
(AK (A"N: CALAN) (CACCA’: AWN AD): 


Le. AK .CA’: AN .A'N. 
Thus 


(cone or segmt. of cone ABB’) : (cone or segmt. of cone APP’) 
= AK CA’ AW AGN. 


ON CONOIDS AND SPHEROIDS. 149 


But (cone or segment of cone APP’) : (segment APP’) 
== As NE” [ Props. 29, 30] 
=e ΝΕ: ΝΕ ΗΝ: 
Therefore, ex aequalt, 
(cone or segment of cone ABB’) : (segment APP’) 
Ξ τ σις PNG INI 


so that (spheroid) : (segment APP’) 
Ξ SAK nd We NY, 
since AA 


Hence (segment A’PP’) : (segment APP’) 
=(HH’.AK—AN.NH'): AN.NH’ 
=(AK.NH+NH’.NK):AN.NH’. 


Further, 
(segment APP’) : (cone or segment of cone APP’) 

ΞΡ]: 71} 
WABI ENE ACN ΠΝ 

and 

(cone or segmt. of cone APP’) : (cone or segmt. of cone A’PP’) 
ΞΡ: ΠΝ 
SABIE SAL Nt AN, 


From the last three proportions we obtain, ex aequale, 
(segment A’PP’) : (cone or segment of cone A’PP’) 
=(AK NAN ONE) < ACN” 
=(AK.NH +N ONE) :(CA’+NH’.CN) 
=(AK.NH+NH .NK):(AK.AN+NH’.CN)...(8). 
But 
AK ΝΗ: AK ANNAN 
=CA+AN:AN 
=AK+CA:CA 


(since AK : AC=AC: AN) 
=HK:CA 


=HK-—-NH:CA-—AN 
=NK:CN 
WEL eee ΝῊ CAN, 


150 ARCHIMEDES. 


Hence the ratio in (8) is equal to the ratio 
ALK ΝΗ AG ΠΝ cor UN ET ΝΕ 


Therefore 
(segment A’PP’) : (cone or segment of cone A’PP’) 
= NIE eA 
=CA+AN: AN. 


[If (ὦ, y) be the coordinates of P referred to the conjugate 
diameters AA’, BB’ as axes of a, y, and if 2a, 2b be the lengths 
of the diameters respectively, we have, since 


(spheroid) — (lesser segment) = ieee segment), 


“ote 


4. ab* — y (a ~«)="" τὴ γῇ (a+2); 


and the above “ is the eet: proof of the truth 


of this equation where a, y are connected by the equation 


co YF 
ates = 1] 


ON SPIRALS. 


“ ARCHIMEDES to Dositheus greeting. 


Of most of the theorems which I sent to Conon, and of 
which you ask me from time to time to send you the proofs, the 
demonstrations are already before you in the books brought to 
you by Heracleides ; and some more are also contained in that 
which I now send you. Do not be surprised at my taking a 
considerable time before publishing these proofs. This has 
been owing to my desire to communicate them first to persons 
engaged in mathematical studies and anxious to investigate 
them. In fact, how many theorems in geometry which have 
seemed at first impracticable are in time successfully worked out! 
Now Conon died before he had sufficient time to investigate 
the theorems referred to; otherwise he would have discovered 
and made manifest all these things, and would have enriched 
geometry by many other discoveries besides. For I know well 
that it was no common ability that he brought to bear on 
mathematics, and that his industry was extraordinary. But, 
though many years have elapsed since Conon’s death, I do not 
find that any one of the problems has been stirred by a single 
person. I wish now to put them in review one by one, 
particularly as it happens that there are two included among 
them which are impossible of realisation* [and which may 
serve as a warning] how those who claim to discover every- 
thing but produce no proofs of the same may be confuted as 
having actually pretended to discover the impossible. 


* Heiberg reads τέλος δὲ ποθεσόμενα, but F has τέλους, so that the true reading 
is perhaps τέλους δὲ ποτιδεόμενα. The meaning appears to be simply ‘ wrong.’ 


152 ARCHIMEDES 


What are the problems I mean, and what are those of which 
you have already received the proofs, and those of which the 
proots are contained in this book respectively, I think it proper 
to specify. The first of the problems was, Given a sphere, to find 
a plane area equal to the surface of the sphere; and this was 
first made manifest on the publication of the book concerning the 
sphere, for, when it is once proved that the surface of any sphere 
is four times the greatest circle in the sphere, it is clear that it 
is possible to find a plane area equal to the surface of the sphere. 
The second was, Given a cone or a cylinder, to find a sphere 
equal to the cone or cylinder; the third, To cut a given sphere 
by a plane so that the segments of it have to one another an 
assigned ratio ; the fourth, To cut a given sphere by a plane so 
that the segments of the surface have to one another an assigned 
- ratio; the fifth, To make a given segment of a sphere similar to 
a given segment of a sphere*; the sixth, Given two segments of 
either the same or different spheres, to find a segment of a sphere 
which shall be similar to one of the segments and have its 
surface equal to the surface of the other segment. The seventh 
was, From a given sphere to cut off a segment by a plane so 
that the segment bears to the cone which has the same base as 
the segment and equal height an assigned ratio greater than 
that of three to two. Of all the propositions just enumerated 
Heracleides brought you the proofs. The proposition stated 
next after these was wrong, viz. that, if a sphere be cut by a 
plane into unequal parts, the greater segment will have to the 
less the duplicate ratio of that which the greater surface has to 
the less. That this is wrong is obvious by what I sent you 
before ; for it included this proposition: If a sphere be cut into 
unequal parts by a plane at right angles to any diameter in the 
sphere, the greater segment of the surface will have to the less 
the same ratio as the greater segment of the diameter has 
to the less, while the greater segment of the sphere has to the 
less a ratio less than the duplicate ratio of that which the 


* τὸ δοθὲν τμᾶμα σφαίρας τῷ δοθέντι τμάματι σφαίρας ὁμοιώσαι, 1.6. to make a 
segment of a sphere similar to one given segment and equal in content to 
another given segment. [Cf. On the Sphere and Cylinder, II. 5.] 


ON SPIRALS. 153 


greater surface has to the less, but greater than the sesqui- 
alterate* of that ratio. The last of the problems was also wrong, 
viz. that, if the diameter of any sphere be cut so that the square 
on the greater segment is triple of the square on the lesser 
segment, and if through the point thus arrived at a plane be 
drawn at right angles to the diameter and cutting the sphere, 
the figure in such a form as is the greater segment of the sphere 
is the greatest of all the segments which have an equal surface. 
That this is wrong is also clear from the theorems which I 
before sent you. For it was there proved that the hemisphere 
is the greatest of all the segments of a sphere bounded by an 
equal surface. 

After these theorems the following were propounded con- 
cerning the conet. If a section of a right-angled cone [a 
parabola], in which the diameter [axis] remains fixed, be made to 
revolve so that the diameter [axis] is the axis [of revolution], 
let the figure described by the section of the right-angled cone 
be called a conoid. And if a plane touch the conoidal figure 
and another plane drawn parallel to the tangent plane cut off 
a segment of the conoid, let the base of the segment cut off be 
defined as the cutting plane, and the vertex as the point in which 
the other plane touches the conoid. Now, if the said figure be 
cut by a plane at right angles to the axis, it is clear that the 
section will be a circle; but it needs to be proved that the 
segment cut off will be half as large again as the cone which has 
the same base as the segment and equal height. And if two 
segments be cut off from the conoid by planes drawn in any 
manner, it is clear that the sections will be sections of acute- 
angled cones [ellipses] if the cutting planes be not at right 
angles to the axis; but it needs to be proved that the 
segments will bear to one another the ratio of the squares on 
the lines drawn from their vertices parallel to the axis to meet 
the cutting planes. The proofs of these propositions are not 
yet sent to you. 

After these came the following propositions about the spiral, 

* (λόγον) μείζονα ἢ ἡμιόλιον τοῦ, ὃν ἔχει κ.τ.λ., i.e. a ratio greater than (the 


ratio of the surfaces) 2, See On the Sphere and Cylinder, 11. 8. 
+ This should be presumably ‘ the conoid,’ not ‘the cone.’ 


154 ARCHIMEDES 


which are as it were another sort of problem having nothing 
in common with the foregoing; and I have written out the 
proofs of them for you in this book. They are as follows. Ifa 
straight line of which one extremity remains fixed be made to 
revolve at a uniform rate in a plane until it returns to the 
position from which it started, and if, at the same time as the 
straight line revolves, a point move at a uniform rate along the 
straight line, starting from the fixed extremity, the point will 
describe a spiral in the plane. I say then that the area 
bounded by the spiral and the straight line which has returned 
to the position from which it started is a third part of the circle 
described with the fixed point as centre and with radius the 
length traversed by the point along the straight line during the 
one revolution. And, if a straight line touch the spiral at the 
extreme end of the spiral, and another straight line be drawn at 
right angles to the line which has revolved and resumed its 
position from the fixed extremity of it, so as to meet the 
tangent, I say that the straight line so drawn to meet it is 
equal to the circumference of the circle. Again, if the revolving 
line and the point moving along it make several revolutions 
and return to the position from which the straight line started, 
I say that the area added by the spiral in the third revolution 
will be double of that added in the second, that in the fourth 
three times, that in the fifth four times, and generally the areas 
added in the later revolutions will be multiples of that added in 
the second revolution according to the successive numbers, 
while the area bounded by the spiral in the first revolution is a 
sixth part of that added in the second revolution. Also, if on 
the spiral described in one revolution two points be taken and 
straight lines be drawn joining them to the fixed extremity of 
the revolving line, and if two circles be drawn with the fixed 
point as centre and radii the lines drawn to the fixed extremity 
of the straight line, and the shorter of the two lines be produced, 
I say that (1) the area bounded by the circumference of the 
greater circle in the direction of (the part of) the spiral included 
between the straight lines, the spiral (itself) and the produced 
straight line will bear to (2) the area bounded by the circum- 
ference of the lesser circle, the same (part of the) spiral and the 


ON SPIRALS. 155 


straight line joining their extremities the ratio which (3) the 
radius of the lesser circle together with two thirds of the excess 
of the radius of the greater circle over the radius of the lesser 
bears to (4) the radius of the lesser circle together with one 
third of the said excess. 


The proofs then of these theorems and others relating to the 
spiral are given in the present book. Prefixed to them, after the 
manner usual in other geometrical works, are the propositions 
necessary to the proofs of them. And here too, as in the books 
previously published, I assume the following lemma, that, if 
there be (two) unequal lines or (two) unequal areas, the excess 
by which the greater exceeds the less can, by being [continually] 
added to itself, be made to exceed any given magnitude among 
those which are comparable with [it and with] one another.” 


Proposition 1. 


If a point move at a uniform rate along any line, and two 
lengths be taken on it, they will be proportional to the times of 
describing them. 


Two unequal lengths are taken on a straight line, and two 
lengths on another straight line representing the times; and 
they are proved to be proportional by taking equimultiples of 
each length and the corresponding time after the manner of 
Euel. V. Def. 5. 


Proposition 2. 


If each of two points on different lines respectively move along 
them each at a uniform rate, and if lengths be taken, one on each 
line, forming pairs, such that each pair are described in equal 
tumes, the lengths will be proportionals. 


This is proved at once by equating the ratio of the lengths 
taken on one line to that of the times of description, which 
must also be equal to the ratio of the lengths taken on the other 
line. 


156 ARCHIMEDES 


Proposition S. 


Given any number of circles, it is possible to find a straight 
line greater than the sum of all their circumferences. 


For we have only to describe polygons about each and then 
take a straight line equal to the sum of the perimeters of the 


polygons. 


Proposition 4. 


Given two unequal lines, viz. a straight line and the circum- 
ference of a circle, it is possible to find a straight line less than 
the greater of the two lines and greater than the less. 


For, by the Lemma, the excess can, by being added a sufficient 
number of times to itself, be made to exceed the lesser line. 


Thus e.g., if ¢ >/ (where ὁ is the circumference of the circle 
and J the length of the straight line), we can find a number n 
such that 

n(c—l)>1. 


Therefore c—l> ᾿ 


and edt >. 


Hence we have only to divide / into n equal parts and add 
one of them tol. The resulting line will satisfy the condition. 


Proposition 5. 


Given a circle with centre O, and the tangent to it at a point 
A, it is possible to draw from O a straight line OPF, meeting the 
circle in P and the tangent in F, such that, if ὁ be the circum- 
Serence of any given circle whatever, 


TP OP <(arecadeyinc: 


Take a straight line, as D, greater than the circumference c. 
[ Prop. 3] 


ON SPIRALS. 157 


Through O draw OH parallel to the given tangent, and 
draw through A a line APH, meeting the circle in P and OH 


A 


er 


Ais 


F 


Ό 


in H, such that the portion PH intercepted between the circle 
and the line OH may be equal to D*. Join OP and produce 
it to meet the tangent in F. 


Then FP: OP=AP: PH, by parallels, 
=AP:D 
<i are ΤΠ} - Ὁ: 


Proposition 6. 


Given a circle with centre O, a chord AB less than the 
diameter, and OM the perpendicular on AB from O, it is possible 
to draw a straight line OF P, meeting the chord AB in F and the 
circle in P, such that 

eis): ΤΣ 
where D: E is any given ratio less than BM : MO. 

Draw OH parallel to AB, and BT perpendicular to BO 
meeting OH in 17. 

Then the triangles BMO, OBT are similar, and therefore 

BM :MO=OB: BT, 
whence Die Os = BT. 


* This construction, which is assumed without any explanation as to how it 
is to be effected, is described in the original Greek thus: ‘‘let PH be placed 
(κείσθω) equal to D, verging (νεύουσα) towards A.”” This is the usual phraseology 
used in the type of problem known by the name of νεῦσις. 


158 ARCHIMEDES 


Suppose that a lme PH (greater than BZ’) is taken such 
that 


D E=0B8 PH, 


E 


and let PH be so placed that it passes through B and P hes on 
the circumference of the circle, while H is on the line OH*. 
(PH will fall outside BT, because PH > BT.) Jom OP meeting 
AB in F. 


We now have 


ie PB — OP Pi 
=O8B PH 
Day, 


Proposition 7. 


Given a circle with centre O, a chord AB less than the 
diameter, and OM the perpendicular on tt from O, it 1s possible 
to draw from O a straight line OPF, meeting the circle in P and 
AB produced in F, such that 


LP PG es 
where D: E is any given ratio greater than BM : MO. 
Draw OT parallel to AB, and BT perpendicular to BO 
meeting OT in 7. 


* The Greek phrase is ‘‘let PH be placed between the circumference and the 
straight line (OH) through B.” The construction is assumed, like the similar 
one in the last proposition. 


ON SPIRALS. 159 


In this case, D:E>BM: MO 
> OB : BT, by similar triangles. 


E 


Take a line PA (less than BT’) such that 
OIG Ja OVER ῬῊ: 
and place PH so that P, H are on the circle and on OT respec- 
tively, while HP produced passes through B*. 
Then Prey OP >) PH 
= 1 I 


Proposition 8. 

Given a circle with centre O, a chord AB less than the 
diameter, the tangent at B, and the perpendicular OM from O 
on AB, it is possible to draw from O a straight line OFP, 
meeting the chord AB in F, the circle in P and the tangent in G, 
such that 

PEG): i, 
where 1): E is any given ratio less than BM : MO. 
If OT be drawn parallel to AB meeting the tangent at B in 7, 
BM: MO=OB: BT, 
so that DE < OB: BT. 
Take a point C on 7B produced such that 
Deve = OB = BC. 
whence BC > BT. 


* PH is described in the Greek as νεύουσαν ἐπὶ (verging to) the point B. As 
before the construction is assumed. 


160 ARCHIMEDES 


Through the points O, 7, C describe a circle, and let OB be 
produced to meet this circle in K. 


Q 


Then, since BC > BT, and OB is perpendicular to OT, it is 
possible to draw from O a straight line OGQ, meeting ΟἽ in G 
and the circle about OTC in Q, such that GQ = Bk*. 


Let OGQ meet AB in F and the original circle in P. 


Now Ca~Gl—0¢G.G0-; 
and OF : 0G=BT: GT, 
so that OF .GT=0G. BT. 


It follows that 
C676 OF. Gl = 0G. G0 20GB, 
or CG: OF = GQ :Br 
= BK: BT, by construction, 
=I Oe 
= BC ΟἹ. 
Hence OP: OF = BC 2 6G. 
and therefore PE OP — EG - δ 
or PE: BG =O0P BC 
= OB: BC 
= Death. 


* The Greek words used are: ‘‘it is possible to place another [straight line] 
GQ equal to KB verging (νεύουσαν) towards O.” This particular νεῦσις is 
discussed by Pappus (p. 298, ed. Hultsch). See the Introduction, chapter νυ. 


ON SPIRALS. 161 


Proposition 9. 


Given a circle with centre O, a chord AB less than the 
diameter, the tangent at B, and the perpendicular OM from O 
on AB, it ts possible to draw from O a straight line OPGF, 
meeting the circle in P, the tangent in G, and AB produced in F, 


such that 
FPYBG=D: E£, 


where D: E is any gwen ratio greater than BM : MO. 


Let OT be drawn parallel to AB meeting the tangent at B 
uae 7. 


Then D:H>BM: MO 
> OB : BT, by similar triangles. 
Produce 78 to C so that 
Dee OB : BC, 
whence BO < BT. 


Describe a circle through the points O, 7, C,and produce OB 
to meet this circle in Κ΄. 


Then, since 7.8.» BC, and OB is perpendicular to CT, it is 
possible to draw from Ὁ a line OGQ, meeting C7 in G, and the 


H. A. HAL 


162 ARCHIMEDES 
circle about OTC in Q, such that GQ= BK*. Let OQ meet 
the original circle in P and AB produced in Ff. 
We now prove, exactly as in the last proposition, that 
CG 70F —BK: Br 


= BC : OP. 
Thus, as before, 
OP :OF=BC : CG, 


and OP =P — BC : BG, 

whence PB Go — OP ΒΟ 
Oa BC, 
ΞΞΙ 7) 


Proposition 10. 


If A,, As, As,...An be n lines forming an ascending arith- 
metical progression in which the common difference is equal 
to A,, the least term, then 

(n+1)Ay?+A,(A,+A.+...+ An) =3 (A+ A+... +Az7). 

[ Archimedes’ proof of this proposition is given above, p. 107— 

9, and it is there pointed out that the result is equivalent to 


γ (9 
1.5.8 Ὁ... ἘΞ ὅ τ 


Cor. 1. It follows from this proposition that 


Ney <o A, + A, +20 Ay )s 
and also that 
ΞΡ. 


[For the proof of the latter inequality see p. 109 8ονϑ. 
Cor. 2. All the results will equally hold if similar figures 


are substituted for squares. 


* See the note on the last proposition. 


ON SPIRALS. 163 


Proposition 11. 


If A,, As,...An be n lines forming an ascending arith- 
metical progression [in which the common difference is equal to 
the least term A,]|*, then 


(n = 1) Hi : (21. a Ana ip de ot a) 
< re : al : A, =F 1 Ge = A,)’} ἢ 
but 
(n — 1) A? : (A aia Dios ΞΡ ooo SF 12) 
>A,’ : {A,.A,+34(An— Aj). 
[Archimedes sets out the terms side by side in the manner 
shown in the figure, where BO=A,, DE = A,_,,...RS = A,, and 
produces DE, FG,...RS until they are 


respectively equal to BC or Ay, so that ¢ y , 
EH, GI,...SU in the figure are re- 
G 
D 


υ 


τ 
[5] 

Μ V7 Sy 
Ρ 


The following proportion is therefore obviously true, viz. 


(n= VAR? G=1)(A,.A, + ΧΆ. ὃ 
AA An tA ALY 


spectively equal to A,, A,...An+. He ἢ 
further measures lengths Bk, DL, 
FM,...PV along BC, DE, FG, ...PQ re- 
spectively each equal to RS. 


The figure makes the relations 
between the terms easier to see with 4 
the eye, but the use of so large a 
number of letters makes the proof 8 
somewhat difficult to follow, and it 
. may be more clearly represented as follows.] 


It is evident that (A, — A,) = Ay. 


= 
“o 


F 


=) 


* The proposition is true even when the common difference is not equal to 
A,, and is assumed in the more general form in Props, 25 and 26. But, as 
Archimedes’ proof assumes the equality of 4: and the common difference, the 
words are here inserted to prevent misapprehension. 


11—2 


164: ARCHIMEDES 
In order therefore to prove the desired result, we have only 
to show that 
(n—1)Ay.A,+3(2—1) Avi’ < (τ +... FA) 
but >(4,.°+4,4 4-24 
I. To prove the first inequality, we have 
(n—1)A,.A,+4(n—1) An” 


=(n—1) A’? +(n—1) A). Anit+}¢(n—1)A,’...(1). 
And 


A, +Anit...+ A, 

= (Anat A,)?+ (Anet+ A)? +... + (414 Ay 
ΞΞ ey te Agee τ eo) 

+(n—1)A,’ 

+2A,(Ayn y+ An ot... +A) 
ξῬ Agee st A,”) 

+(n—1) A; 

+A, {AnitAn+et+Anst... +A 

+A, +A, +... +An2+ Ano} 

Ξ απ 1.9 

+(n—1) A,’ 

MAL Wa od, « iallaies oeid aerate hohe a eee (2). 


Comparing the right-hand sides of (1) and (2), we see that 
(n—1)A,’ is common to both sides, and 


(W—1) A.;, Ayo dole 
while, by Prop. 10, Cor. 1, 


4(n—1) An? < Any’ +Ana't+... + Ay. 
It follows therefore that 
(n=1)A,- Aled — DA, Ξ ae eee 
and hence the first part of the proposition is proved. 


II. We have now, in order to prove the second result, to 
show that 


(n—1)A,.A,+4(n—1) Ay? > (An? + An’ +... +A4,°). 


ON SPIRALS. 165 


The right-hand side is equal to 
(Anot A, + (An s+ Ai)? +...+(4,+4,7+4/ 

=A, +Ans +... FA? 
+(n-—1)A,?? 
+2A,(AnotAnst... +A,) 

(Ayo baa ta A) 
+(n—1)A? 
caps AnstAnst...+A; 

AG As τῷ b Ane 

= (An? + Ans +... + 4.2 
+(n—1)A/? 
ἘΞ Vee Ay, προ οοόο ate velco πὺὰν (3). 


Comparing this expression with the right-hand side of (1) above, 
we see that (n—1) A,’ is common to both sides, and 


(n—1)A,.Ayn.>(n—2)A,. Ano, 
while, by Prop. 10, Cor. 1, 
4 (n—1) Api? > (An + Ane’ +... +A’). 
Hence 
(n—1)A,.A,+3(n—1) An? > (An? + An’ +... + 40; 

and the second required result follows. 

Cor. The results in the above proposition are equally true if 
similar figures be substituted for squares on the several lines. 

DEFINITIONS. 


1. Ifa straight line drawn in a plane revolve at a uniform 
rate about one extremity which remains fixed and return to 
the position from which it started, and if, at the same time as 
the line revolves, a point move at a uniform rate along the 
straight line beginning from the extremity which remains fixed, 
the point will describe a spiral (ἕλιξ) in the plane. 


2. Let the extremity of the straight line which remains 


166 ARCHIMEDES 


fixed while the straight line revolves be called the origin* 
(apya) of the spiral. 


3. And let the position of the line from which the straight 
line began to revolve be called the initial line* in the 
revolution (apya τᾶς περιφορᾶς). 


4. Let the length which the point that moves along the 
straight line describes in one revolution be called the first 
distance, that which the same point describes in the second 
revolution the second distance, and similarly let the distances 
described in further revolutions be called after the number of 
the particular revolution. 


5. Let the area bounded by the spiral described in the 
first revolution and the first distance be called the first area, 
that bounded by the spiral described in the second revolution 
and the second distance the second area, and similarly for the 
rest in order. 


6. If from the origin of the spiral any straight line be 
drawn, let that side of it which is in the same direction as that 
of the revolution be called forward (spoayovpeva), and that 
which is in the other direction backward (ἑπόμενα). 


7. Let the circle drawn with the origin as centre and the 
Jirst distance as radius be called the first circle, that drawn 
with the same centre and twice the radius the second circle, 
and similarly for the succeeding circles. 


Proposition 12. 


If any number of straight lines drawn from the origin to 
meet the spiral make equal angles with one another, the lines will 
be in arithmetical progression. 


[The proof is obvious. ] 
* The literal translation would of course be the “beginning of the spiral” 


and ‘“‘the beginning of the revolution” respectively. But the modern names 
will be more suitable for use later on, and are therefore employed here. 


ON SPIRALS. 167 


Proposition 13. 
Tf a straight line touch the spiral, τέ will touch it in one point 
only. 
Let O be the origin of the spiral, and BC a tangent to it. 


If possible, let BC touch the spiral in two points P, Q. 
Join OP, OQ, and bisect the angle POQ by the straight line OR 
meeting the spiral in R. 


Then [Prop. 12] OR is an arithmetic mean between OP and 
OQ, or 
OP + 0Q = 208. 
But in any triangle POQ, if the bisector of the angle POQ 
meets PQ in K, 
OP + 0Q > 20K *. 
Therefore OX < OR, and it follows that some point on BC 
between P and Q lies within the spiral. Hence BC cuts the 
spiral; which is contrary to the hypothesis. 


Proposition 14. 


If O be the origin, and P, Q two points on the first turn of 
the spiral, and if OP, OQ produced meet the ‘first circle’ 
AKP'(’ in P’, Q’ respectively, OA being the initial line, then 

OP ἡ — (are AICP ): Gre 4 KO’). 

For, while the revolving line OA moves about O, the point 

A on it moves uniformly along the circumference of the circle 


* This is assumed as a known proposition ; but it is easily proved. 


168 ARCHIMEDES 


AKP’Q’, and at the same time the point describing the spiral 
moves uniformly along OA. 


Ρ' Q’ 


ca, 


K 


Thus, while A describes the arc AXP’, the moving point on 
OA describes the length OP, and, while A describes the are 
AKQ’, the moving point on OA describes the distance OQ. 


Hence OP: 0Q=(are AKP’): (arc AKQ’). _— [Prop. 2] 


Proposition 15. 


If P, Q be points on the second turn of the spiral, and OP, 
OQ meet the ‘first circle’ AKP’Q’ in P’, Q’, as in the last 
proposition, and if c be the circumference of the first circle, then 


OP : OQ=c+¢ (arc AKP’) : c+ (arc AKQ’). 


For, while the moving point on OA describes the distance 
OP, the point A describes the whole of the circumference of 
the ‘first circle’ together with the are AKP’; and, while the 
moving point on OA describes the distance OQ, the point A 
describes the whole circumference of the ‘first circle’ together 
with the are AKQ’. 


Cor. Similarly, if P, Q are on the nth turn of the spiral, 
OP : OQ=(n—1)c+(are AKP’): (κ -- 1) 6- (τὸ AKQ’). 


ON SPIRALS. 169 


Propositions 16, 17. 


If BC be the tangent at P, any point on the spiral, PC being 
the ‘ forward’ part of BC, and if OP be joined, the angle OPC 
as obtuse while the angle OPB is acute. 


I. Suppose P to be on the first turn of the spiral. 

Let OA be the initial line, AK P’ the ‘first circle. Draw 
the circle DEP with centre O and radius OP, meeting OA in 
D. This circle must then, in the ‘forward’ direction from P, 


fall within the spiral, and in the ‘backward’ direction outside 
it, since the radii vectores of the spiral are on the ‘ forward ’ side 
greater, and on the ‘ backward’ side less, than OP. Hence the 
angle OPC cannot be acute, since it cannot be less than the 
angle between OP and the tangent to the circle at P, which is 
a right angle. 


It only remains therefore to prove that OPC is not a right 
angle. 


If possible, let it be a right angle. BC will then touch 
the circle at P. 


Therefore [Prop. 5] it is possible to draw a line OQC 
meeting the circle through P in Q and BC in ΟἹ such that 


CO OO στο PO) (are ΡΒ... aly 


170 ARCHIMEDES 


Suppose that OC meets the spiral in R and the ‘first circle’ 
in R’; and produce OP to meet the ‘first circle’ in P’. 


From (1) it follows, componendo, that 
CO : OQ < (are DLQ) : (arc DLP) 
<(are AR’): (are AKE) 
=O OP; [Prop. 14] 
But this is impossible, because OQ = OP, and OR < OC. 


Hence the angle OPC is not a right angle. It was also 
proved not to be acute. 


Therefore the angle OPC is obtuse, and the angle OPB 
consequently acute. 


II. If P is on the second, or the nth turn, the proof is the 
same, except that in the proportion (1) above we have to 
substitute for the are DEP an are equal to (p+are DLP) or 
(n—1.p+are DLP), where p is the perimeter of the circle 


DLP through P. Similarly, in the later steps, p or (n—1) p 
will be added to each of the arcs DLQ and DLIP, and ὁ or 
(n—1)c to each of the arcs AKR’, AKP’, where ὁ is the 
circumference of the ‘first circle’ AKP’. 


ON SPIRALS. 171 


Propositions 18, 19. 


I. If OA be the initial line, A the end of the first turn of 
the spiral, and if the tangent to the spiral at A be drawn, the 
straight line OB drawn from O perpendicular to OA will meet 
the said tangent in some point B, and OB will be equal to the 
corcumference of the ‘ first circle. 


Il. If A’ be the end of the second turn, the perpendicular 
OB will meet the tangent at A’ in some point B’, and OB’ will 
be equal to 2 (circumference of ‘ second circle’). 


Ill. Generally, if A, be the end of the nth turn, and OB 
meet the tangent at A, in B,, then 


OBA= Cp, 


where Cn 18 the circumference of the ‘nth circle.’ 


I. Let AKC be the ‘first circle. Then, since the ‘ back- 
ward’ angle between OA and the tangent at A is acute [Prop. 
16], the tangent will meet the ‘first circle’ in a second point C. 
And the angles CAO, BOA are together less than two right 
angles; therefore OB will meet AC produced in some point B. 


Then, if ὁ be the circumference of the first circle, we have 
to prove that 


OB =c. 


If not, OB must be either greater or less than c. 


Le ARCHIMEDES 


(1) If possible, suppose OB >. 
Measure along OB a length OD less than OB but greater 
than c. 


We have then a circle AKC,a chord AC in it less than 
the diameter, and a ratio AO: OD which is greater than the 
ratio AO: OB or (what is, by similar triangles, equal to it) the 
ratio of 146 to the perpendicular from Ὁ on AC. Therefore 
[Prop. 7] we can draw a straight line OPF, meeting the circle 
in P and CA produced in F, such that 


PE! PA=AO : OD, 
Thus, alternately, since AO = PO, 
Teo PA Op 
<(are PA) τ 
since (arc PA) > PA, and OD >c. 


Componendo, 
FO? PO <(¢-- are PA) 3 
- 00204, 
where OF meets the spiral in Q. [Prop. 15] 
Therefore, since 04 = OP, FO < OQ; which is impossible. 
Hence OB +. 


(2) If possible, suppose OB < ο. 

Measure OF along OB so that OF is greater than OB but 
less than c. 

In this case, since the ratio AO: OF is less than the ratio 
AO: OB (or the ratio of 46 to the perpendicular from Ὁ 
on AC), we can [Prop. 8] draw a line OF" P’G, meeting AC in 
F’, the circle in P’, and the tangent at A to the circle in G, 
such that 

FOP TA'G= AQ: OF. 
Let OP'G cut the spiral in Q’. 
Then we have, alternately, 
ee — AG OE, 
> (arc ΠΕ: Ὁ: 
because AG > (are AP’), and OF « ο. 


ON SPIRALS. i aye) 


Therefore 
POE OU —a(are AK ΤΟ}: ὁ 
< O00 : OA. [Prop. 14] 
But this is impossible, since OA = OP’, and OQ’ < OF". 
Hence OB 4.6. 
Since therefore OB is neither greater nor less than ὁ, 
OB — cc: 


II. Let A’K’O’ be the ‘second circle, A’C’ being the 
tangent to the spiral at A’ (which will cut the second circle, 
since the ‘backward’ angle OA’C’ is acute). Thus, as before, 
the perpendicular OB’ to OA’ will meet A’C’ produced in some 
point B’. 

If then οἱ is the circumference of the ‘second circle,’ we 
have to prove that OB’ = 2c’. 


For, if not, OB’ must be either greater or less than 2c’. 
(1) If possible, suppose OB’ > 2c’. 


Measure OD’ along OB’ so that OD’ is less than OB’ but 
greater than 2c’. 


Then, as in the case of the < first circle’ above, we can draw 
a straight line OPF' meeting the ‘second circle’ in P and C’A’ 
produced in F, such that 


HP: PAE Ae OD. 


174 ARCHIMEDES 


Let OF meet the spiral in Q. 
We now have, since A’O = PO, 
iP Oey 0! 
<‘(are A’P):2c7, 
because (arc A’P) > A’P and OD’ > 26. 
Therefore FO: PO 2@e¢ 4 arc A’P) = 2¢ 


00 OA’. [Prop. 15, Cor.] 
Hence FO < OQ; which is impossible. 
Thus OB’ > 26. 
Similarly, as in the case of the ‘ first circle’, we can prove that 
OB’ €< 2c’. 
Therefore OE 2c. 


III. Proceeding, in like manner, to the ‘third’ and suc- 
ceeding circles, we shall prove that 


OB, = hops 


Proposition 20. 


I. 17 P be any point on the first turn of the spiral and OT 
be drawn perpendicular to OP, OT will meet the tangent at P to 
the spiral in some point T'; and, if the circle drawn with centre 
O and radius OP meet the initial line in K, then OT 1s equal to 
the arc of this circle between K and P measured in the ‘forward’ 
direction of the spiral. 


11. Generally, of P be a point on the nth turn, and the 
notation be as before, while p represents the circumference of the 
circle with radius OP, 


OT =(n—1)p+are KP (measured ‘ forward’). 
I. Let P be a point on the first turn of the spiral, OA the 
initial line, PR the tangent at P taken in the ‘backward’ 
direction. 


Then [Prop. 16] the angle OPR is acute. Therefore Ph 


ON SPIRALS. 175 
meets the circle through P in some point #&; and also OT will 
meet PR produced in some point 7. 


If now OT is not equal to the arc KAP, it must be either 
greater or less. 


A 


(1) If possible, let OT be greater than the are XH RP. 

Measure OU along OT less than OT but greater than the 
arc KRP. 

Then, since the ratio PO: OU is greater than the ratio 
PO: OT, or (what is, by similar triangles, equal to it) the 
ratio of $PR to the perpendicular from Ὁ on PR, we can draw 
a line OQF, meeting the circle in Q and RP produced in F, 
such that 

ROP PhO OU. [ Prop. 7] 
Let OF meet the spiral in Q’. 
We have then 
FQ -Q0— PQ - 00 
< (are PQ) : (are KRP), by hypothesis. 
Componendo, 
FO : QO < (are KRQ) : (arc KRP) 


< 00 OP. [Prop. 14] 
But QO = OP. 


176 ARCHIMEDES 


Therefore FO < ΟΘ΄; which is impossible. 
Hence OT + (are KRP). 


(2) The proof that O7 + (are K RP) follows the method of 
Prop. 18, I. (2), exactly as the above follows that of Prop. 18, 
Ta). 

Since then O7' is neither greater nor less than the are KAP, 
it is equal to it. 


1. If P be on the second turn, the same method shows 
that 
OT =p+(arc KRP); 
and, similarly, we have, for a point P on the nth turn, 


OT =(n—1)p+(are KRP). 


Propositions 21, 22, 29. 


Given an area bounded by any arc of a spiral and the lines 
joining the extremities of the arc to the origin, it 1s possible to 
circumscribe about the area one figure, and to inscribe in tt 
another figure, each consisting of similar sectors of circles, and 
such that the circumscribed figure exceeds the inscribed by less 
than any assigned area. 


For let BC be any arc of the spiral, O the origin. Draw 
the circle with centre O and radius OC, where C is the ‘forward’ 
end of the are. 


Then, by bisecting the angle BOC, bisecting the resulting 
angles, and so on continually, we shall ultimately arrive at 
an angle COr cutting off a sector of the circle less than any 
assigned area. Let COr be this sector. 


Let the other lines dividing the angle BOC into equal parts 
meet the spiral in P, Q, and let Or meet it in &. With O as 
centre and radii OB, OP, OQ, OF respectively describe arcs of 
circles Bp’, bBq’, pQr’, qRc', each meeting the adjacent radii as 
shown in the figure. In each case the are in the ‘forward’ 
direction from each point will fall within, and the arc in the 
‘backward’ direction outside, the spiral. 


ON SPIRALS. 177 


We have now a circumscribed figure and an inscribed figure 
each consisting of similar sectors of circles. To compare their 
areas, we take the successive sectors of each, beginning from OC, 
and compare them. 


The sector OCr in the circumscribed figure stands alone. 
And (sector ORq) = (sector ORc’), 
(sector OQp) = (sector OQr’), 
(sector OPb) = (sector OP’), 
while the sector OBp’ in the inscribed figure stands alone. 


Hence, if the equal sectors be taken away, the difference be- 
tween the circumscribed and inscribed figures is equal to the 
difference between the sectors OCr and OBp’; and this difference 
is less than the sector OCr, which is itself less than any 
assigned area. 

The proof is exactly the same whatever be the number of 
angles into which the angle BOC is 
divided, the only difference being 
that, when the arc begins from the 
origin, the smallest sectors OPb, OPq’ 
in each figure are equal, and there is 
therefore no inscribed sector standing 
by itself, so that the difference 
between the circumscribed and in- 
scribed figures is equal to the sector 
OCr itself. 


H. A. 12 


178 ARCHIMEDES 


Thus the proposition is universally true. 


Cor. Since the area bounded by the spiral is intermediate 
in magnitude between the circumscribed and inscribed figures, 
it follows that 

(1) ὦ figure can be circumscribed to the area such that it 
exceeds the area by less than any assigned space, 

(2) a figure can be inscribed such that the area exceeds it by 
less than any assigned space. 


Proposition 24. 


The area bounded by the first turn of the spiral and the 
initial line is equal to one-third of the ‘first circle’ [- π᾿ (27a)’, 
where the spiral is r=a@]. 

[The same proof shows equally that, if OP be any radius 
vector in the first turn of the spiral, the area of the portion of 
the spiral bounded thereby ts equal to one-third of that sector of 
the circle drawn with radius OP which is bounded by the intial 
line and OP, measured in the ‘forward’ direction from the 
unitial line. | 

Let O be the origin, OA the initial line, A the extremity of 
the first turn. 


Draw the ‘first circle, 1.6, the circle with O as centre and 
OA as radius. 
Then, if C, be the area of the first circle, R, that of the first 
turn of the spiral bounded by OA, we have to prove that 
R, = 4C;. 


For, if not, R, must be either greater or less than C}. 


I. If possible, suppose A, < 10]. 
We can then circumscribe a figure about R, made up of 
similar sectors of circles such that, if / be the area of this 


figure, 
F-R,<1¢C,-R8,, 
whence F< 10]. 


ON SPIRALS. 179 


Let OP, OQ, ... be the radii of the circular sectors, beginning 
from the smallest. The radius of the largest is of course OA. 


The radii then form an ascending arithmetical progression 
in which the common difference is equal to the least term OP. 
If n be the number of the sectors, we have [by Prop. 10, Cor. 1] 


n.0A?<3(OP* + 00 +...+ 0A’); 


and, since the similar sectors are proportional to the squares on 
their radii, it follows that 


C, < 3F, 

or F>1C,. 
But this is impossible, since F' was less than 1C,. 

Therefore R, € £0). 


II. If possible, suppose R, > 16. 


We can then inscribe a figure made up of similar sectors of 

circles such that, if f be its area, 
R,-f< Rk, -—1C, 

whence f> τὰ 

If there are (n — 1) sectors, their radii, as OP, OQ,..., form 
an ascending arithmetical progression in which the least term 
is equal to the common difference, and the greatest term, as 
OY, is equal to (n -- 1) OP. 

12—2 


180 ARCHIMEDES 


Thus [Prop. 10, Cor. 1] 
n.OA?>38(OP? + 0Q’?+...+ ΟΥ̓", 


whence σι > 3f, 
or f<iC,; 
which is impossible, since f> 10]. 
Therefore R, +1C,. 
Since then R, is neither greater nor less than 4C,, 
ΞΟ 


[Archimedes does not actually find the area of the spiral 
cut off by the radius vector OP, where P is any point on the 
first turn; but, in order to do this, we have only to substitute 


Ρ 


in the above proof the area of the sector KLP of the circle 
drawn with O as centre and OP as radius for the area C, of 
the ‘ first circle’, while the two figures made up of similar sectors 
have to be circumscribed about and inscribed in the portion 
OEP of the spiral. The same method of proof then applies 
exactly, and the area of OEP is seen to be 4 (sector KLIP). 


We can prove also, by the same method, that, if P be a 
point on the second, or any later turn, as the nth, the complete 
area described by the radius vector from the beginning up to 
the time when it reaches the position OP is, if C denote the 
area of the complete circle with O as centre and OP as radius, 
1 (C+ sector KLP) or } (n—1.C+ sector KLIP) respectively. 


The area so described by the radius vector is of course not 
the same thing as the area bounded by the last complete turn 


ON SPIRALS. 181 


of the spiral ending at P and the intercepted portion of the 
radius vector OP. Thus, suppose A, to be the area bounded 
by the first turn of the spiral and OA, (the first turn ending at 
A, on the initial line), A, the area added to this by the second 
complete turn ending at A, on the initial line,and so on. R, has 
then been described twice by the radius vector when it arrives 
at the position OA,; when the radius vector arrives at the 
position OA,, it has described R, three times, the ring R, twice, 
and the ring R; once; and so on. 


Thus, generally, if C,, denote the area of the ‘nth circle, we 
shall have 
inC,, = R, + ὯΝ, + 3R,.+... +R, 
while the actual area bounded by the outside, or the complete 
nth, turn and the intercepted portion of OA, will be equal to 


Rk, — Ry ah Ry-s Spa Con R,. 


It can now be seen that the results of the later Props. 25 
and 26 may be obtained from the extension of Prop. 24 just 
given. 


To obtain the general result of Prop. 26, suppose BC to be 
an arc on any turn whatever of the spiral, being itself less than 
a complete turn, and suppose B to be beyond A, the extremity 
of the nth complete turn, while C is ‘forward’ from B. 


Let - be the fraction of a turn between the end of the nth 
turn and the point B. 


Then the area described by the radius vector up to the 
position OB (starting from the beginning of the spiral) is 
equal to 


i (n =) (circle with rad. OB), 


Also the area described by the radius vector from the beginning 
up to the position OC is 


᾿ ἵ n = (circle with rad. OC) + (sector B’M 6) ; 


182 ARCHIMEDES 


The area bounded by OB, OC and the portion BEC of the 
spiral is equal to the difference between these two expressions ; 
and, since the circles are to one another as OB’ to OC’, the 
difference may be expressed as 


2 
4 ἴι =) (1 = aa) (circle with rad. OC) + (sector Β΄ ΗΠ o| ; 


But, by Prop. 15, Cor., 
(ι τ (circle B’MC) : \(n τ (circle B’MC) + (sector B’M 0)| 


=OB: OC, 


so that 
(n +") (circle B’MC) : (sector B’MC) = OB : (OC — OB). 


Thus 3188 BEG) oo. 1( OB ) (1 ocr) 1} 


sector BMC *\\OC—OB/\ OC? 
_ , OB(OC + OB) + 00" 
mes 06: 
_ 0C.0B+4(0C — OBY 
Ῥῇ OC? 


The result of Prop. 25 is a particular case of this, and the 
result of Prop. 27 follows immediately, as shown under that 
proposition. ] 


ON SPIRALS. 183 


Propositions 25, 26, 27. 


[Prop. 25.] If A, be the end of the second turn of the spiral, 
the area bounded by the second turn and OA, is to the area 
of the ‘second circle’ in the ratio of 7 to 12, being the ratio of 
{ry +4(72—1))} to ry’, where γι, r, are the radia of the ‘ first’ 
and ‘ second’ circles respectively. 

[Prop. 26.] Jf BC be any arc measured in the ‘forward’ 
direction on any turn of a spiral, not being greater than the 
complete turn, and tf a circle be drawn with O as centre and OC 
as radius meeting OB in B’, then 

(area of spiral between OB, OC) : (sector OB’C) 
= {0C.0B+4(00— OBY} : OC*. 

[Prop. 27.] If R, be the area of the first turn of the spiral 
bounded by the initial line, R, the area of the ring added by the 
second complete turn, R; that of the ring added by the third turn, 
and so on, then 

ig Aisa oe ie 2 hoe (7 — eas 
Also i= 6h: 

[Archimedes’ proof of Prop. 25 is, mutatis mutandis, the 
same as his proof of the more general Prop. 26. The latter 
will accordingly be given here, and applied to Prop. 25 as a 
particular case. ] 

Let BC be an arc measured in the ‘forward’ direction on 
any turn of the spiral, CK B’ the circle drawn with O as centre 
and OC as radius. 

Take a circle such that the square of its radius is equal 
to OC.O0B+4(OC— OB)’, and let o be a sector in it whose 
central angle is equal to the angle BOC. 

Thus σ΄: (sector OB'C) = {0OC. OB + 4(0C — OB)*} : OC, 
and we have therefore to prove that 


(area of spiral OBC) = c. 


For, if not, the area of the spiral OBC (which we will call S) 
must be either greater or less than o. 


184 ARCHIMEDES 


I. Suppose, if possible, S < σ. 

Circumscribe to the area S a figure made up of similar 
sectors of circles, such that, if F be the area of the figure, 

F-—S<a-S, 

whence To: 

Let the radii of the successive sectors, starting from OB, 
be OP, OQ,...0C. Produce OP, OQ,... to meet the circle 
CECB 3x. 


If then the lines OB, OP, OQ,... OC be n in number, the 
number of sectors in the circumscribed figure will be (n — 1), 
and the sector OB’C will also be divided into (n—1) equal 
sectors. Also OB, OP, OQ,...OC will form an ascending 


arithmetical progression of n terms. 
Therefore [see Prop. 11 and Cor.] 
(n—1) OC? : (OP? + 0Q? +... + OC") 
< ΟΟ" : {0C. OB + 4(0C — OB)*} 
< (sector OB’C) : σ, by hypothesis. 
Hence, since similar sectors are as the squares of their radii, 
(sector OB’C) : F< (sector OB’C) : a, 
so that Fi >a. 
But this is impossible, because F’ < σ. 
Therefore Sto. 


ON SPIRALS. 185 


II. Suppose, if possible, S Ὁ σ. 
Inscribe in the area S a figure made up of similar sectors of 
circles such that, if f be its area, 
5 -- "i «ἢ -- σ, 
whence pa 
Suppose OB, OP,...OY to be the radii of the successive 
sectors making up the figure f, being (nm — 1) in number. 
We shall have in this case [see Prop. 11 and Cor.] 
(n—1) OC? : (OB? + OP’ +...+ OY”) 
> OC’: {OC.OB + 4(00 — OB)’}, 
whence (sector OB’C) : f > (sector OB’C) : a, 
so that (een 
But this is impossible, because f > σ. 
Therefore Sto. 
Since then S is neither greater nor less than a, it follows that 
ΞΘ 
In the particular case where B coincides with A,, the end 
of the first turn of the spiral, and C with A,, the end of the 
second turn, the sector OB’C becomes the complete ‘second 
circle, that, namely, with OA, (or 7.) as radius. 
Thus 
(area of spiral bounded by OA.) : (‘second circle’) 
={rnit+i—n)} i 
=(2+4):4 (since r,= 27) 
ΞΞ 51 
Again, the area of the spiral bounded by OA, is equal to 
R,+ R, (i.e. the area bounded by the first turn and OA,, 
together with the ring added by the second turn). Also the 


‘second circle’ is four times the ‘first circle, and therefore 
equal to 12 R,. 


Hence Ceri) hae Ξ τ 
or R, + Ra = th. 
Thus JT oa eee: Be CL): 


2 


186 ARCHIMEDES 
Next, for the third turn, we have 
(R, + R,+ R;) : (‘third circle’) = {727 + 4 (1s — 72} : re 
=(§.244)73° 


= a. 

and (‘third circle’) = 9 (‘first circle’) 

= “idee 
therefore i, By +R, = 19:2, 
and, by (1) above, it follows that 

el Aes 

= Dil by vase \eones0 yiagoheeeeeeeme (2) 

and so on. 


Generally, we have 
(R, + Bo+...+R,): (nth circle) = {rarn + 4 (Ta — Tray} i Tn’ 
ΓΕ ee eth circle) 
=a Ta— +t Ta = Tn) | ee 
and (nth circle) : (n — 1th Citcle) —97, tat: 
Therefore 
(2, + R,+...+R,): (Ri + B+... + Bar) 
= {n(n- 1) +4}: (@-1)(m-2) 48 
= {8n(n—1)+1}: (8 (υ -- 1)(ι -- 2) + 1). 


Dirimendo, 
R,: (Rh, + Ro +... + Rr) 

=6(n—1): {83(n—1)(n—2) 4-1} «ννννν νον (a). 
Similarly 


Raya: (Ri + Re +... + Rn») = 6 (n — 2) : (8 (ὦ — 2) (n — 8) + 1), 
from which we derive 
Rage τ Pha cb nas Ὁ 1...) 
= 6(n—2): {6(n—2)+3(n—2)(n—3)+4+]1} 
= 6(n—2): {8(n—1)(n—2) 4 IT}... (8). 


ON SPIRALS. 187 


Combining (a) and (8), we obtain 
Ry: Raa =(n—1): (n—- 2). 
Thus 


R,, R;, R,y,... Ry are in the ratio of the successive numbers 
1,2,3...(n—1). 


Proposition 28. 


If O be the origin and BC any arc measured in the ‘forward’ 
direction on any turn of the spiral, let two circles be drawn 
(1) with centre O, and radius OB, meeting OC in C’, and 
(2) with centre O and radius OC, meeting OB produced in Β΄. 
Then, if E' denote the area bounded by the larger circular arc 
B’C, the line B’B, and the spiral BC, while F denotes the area 
bounded by the smaller arc BC’, the line CC’ and the spiral BC, 

EH: F={0B+2(0C — OB)} : {OB -- ὦτ (006 — OB). 


Let o denote the area of the lesser sector OBC’; then the 
larger sector OB’C is equal too + F+ EF. 


Thus [Prop. 26] 
(o+ Ff) :(¢+F+ £)={00.0B +4 (OC — OB) : OC’...(1), 
whence 
E :(o + F) = {0C(0C -- OB) -- 4(0C — OB 
: {OC.O0B+4(0C — OB} 
= {OB (06 — OB) + 2(0C — OBY} 
: {0C.OB + 4(OC — OB)Y}............ (2). 


188 ARCHIMEDES ON SPIRALS. 
Again 
(c+ F+#):c0=0C": OB’. 
Therefore, by the first proportion above, ea aequali, 
(c+ F):¢={0C0.0B+4(0C — OB)’} : 08", 
whence 
(¢+ fF): F={0C.0B +1(0C — OB)’} 
: {OB(OC — OB) + (00 — OB)’. 
Combining this with (2) above, we obtain 
E : F={0B(0C — OB) + 2 (OC — OB) 
: {OB (OC — OB) + 4(0C — OB)’} 
= {OB +2(0C — OB)} : {OB + τ(06 — OB)}. 


ON THE EQUILIBRIUM OF PLANES 
OR 


THE CENTRES OF GRAVITY OF PLANES. 


BOOK 1. 


“JT POSTULATE the following: 


1. Equal weights at equal distances are in equilibrium, 
and equal weights at unequal distances are not in equilibrium 
but incline towards the weight which is at the greater distance. 


2. If, when weights at certain distances are in equilibrium, 
something be added to one of the weights, they are not in 
equilibrium but incline towards that weight to which the 
addition was made. 


3. Similarly, if anything be taken away from one of the 
weights, they are not in equilibrium but incline towards the 
weight from which nothing was taken. 


4. When equal and similar plane figures coincide if applied 
to one another, their centres of gravity similarly coincide. 


5. In figures which are unequal but similar the centres of 
gravity will be similarly situated. By points similarly situated 
in relation to similar figures I mean points such that, if straight 
lines be drawn from them to the equal angles, they make equal 
angles with the corresponding sides. 


190 ARCHIMEDES 


6. If magnitudes at certain distances be in equilibrium, 
(other) magnitudes equal to them will also be in equilibrium at 
the same distances. 

7. In any figure whose perimeter is concave in (one and) 
the same direction the centre of gravity must be within the 


figure.” 


Proposition 1. 


Weights which balance at equal distances are equal. 

For, if they are unequal, take away from the greater the 
difference between the two. The remainders will then not 
balance [Post. 3]; which is absurd. 

Therefore the weights cannot be unequal. 


Proposition 2. 


Unequal weights at equal distances will not balance but will 
incline towards the greater weight. 

For take away from the greater the difference between the 
two. The equal remainders will therefore balance [Post. 1]. 
Hence, if we add the difference again, the weights will not 
balance but incline towards the greater [Post. 2]. 


Proposition S. 
Unequal weights will balance at unequal distances, the greater 
weight being at the lesser distance. 


Let A, B be two unequal weights (of which A is the 
greater) balancing about ( at distances AC, BC respectively. 


Then shall AC be less than BC. For, if not, take away 
from A the weight (4 --.8.)ὺ The remainders will then incline 


ON THE EQUILIBRIUM OF PLANES. I. 191 


towards B [ Post. 3]. But this is impossible, for (1) if AC= CB, 
the equal remainders will balance, or (2) if AC > CB, they will 
incline towards A at the greater distance [Post. 1]. 


Hence AC < CB. 


Conversely, if the weights balance, and AC < OB, then 
A>bB. 


Proposition 4. 


If two equal weights have not the same centre of gravity, the 
centre of gravity of both taken together is at the middle point of 
the line jowning their centres of gravity. 


[Proved from Prop. 3 by reductio ad absurdum. Archimedes 
assumes that the centre of gravity of both together is on the 
straight line joining the centres of gravity of each, saying that 
this had been proved before (προδέδεικται). The allusion is no 
doubt to the lost treatise On levers (περὶ ζυγῶν).] 


Proposition 5. 


If three equal magnitudes have their centres of gravity on a 
straight line at equal distances, the centre of gravity of the 
system will coincide with that of the middle magnitude. 


[This follows immediately from Prop. 4.] 


Cor 1. The same is true of any odd number of magnitudes 
uf those which are at equal distances from the middle one are 
equal, while the distances between their centres of gravity are 
equal. 


Cor. 2. If there be an even number of magnitudes with 
their centres of gravity situated at equal distances on one straight 
line, and vf the two middle ones be equal, while those which are 
equidistant from them (on each side) are equal respectively, the 
centre of gravity of the system 1s the middle point of the line 
joining the centres of gravity of the two middle ones. 


192 ARCHIMEDES 


Propositions 6, 7. 


Two magnitudes, whether commensurable [Prop. 6] or %n- 
commensurable [Prop. 7], balance at distances reciprocally 
proportional to the magnitudes. 


I. Suppose the magnitudes A, B to be commensurable, 
and the points A, B to be their centres of gravity. Let DH be 
a straight line so divided at C that 

Ai B=DC: CE. 


We have then to prove that, if A be placed at # and B at 
D, Cis the centre of gravity of the two taken together. 


τ 


Ν 


Since A, B are commensurable, so are DC, CE. Let N be 
a common measure of DO, CH. Make DH, DK each equal to 
CE, and EL (on CE produced) equal to CD. Then HH=CD, 
since DH=CEH. Therefore LH is bisected at Δ᾽, as HK is 
bisected at D. 


Thus LH, HK must each contain V an even number of 
times. 


Take a magnitude O such that O is contained as many 
times in A as J is contained in LH, whence 


A: O= TT ay: 
But B:A=CE : DC 
= He Tid. 


Hence, ex aequali, B: O=HK: N, or O is contained in B as 
many times as JV is contained in HK. 


Thus O is a common measure of A, B. 


ON THE EQUILIBRIUM OF PLANES I. 193 


Divide LH, HK into parts each equal to V, and A, B into 
parts each equal to 0. The parts of A will therefore be equal 
in number to those of LH, and the parts of B equal in number 
to those of HK. Place one of the parts of A at the middle 
point of each of the parts V of LH, and one of the parts of B 
at the middle point of each of the parts V of HK. 

Then the centre of gravity of the parts of A placed at equal 
distances on LH will be at #, the middle point of LH [Prop. 5, 
Cor. 2], and the centre of gravity of the parts of B placed at 
equal distances along HK will be at D, the middle point of HK. 

Thus we may suppose A itself applied at H, and B itself 
applied at D. 

But the system formed by the parts O of A and B together 
is a system of equal magnitudes even in number and placed at 
equal distances along LK. And, since LE = CD, and EC= DK, 
LC= CK, so that C is the middle point of LK. Therefore C is 
the centre of gravity of the system ranged along LK. 

Therefore A acting at H and B acting at D balance about 
the point C. 


II. Suppose the magnitudes to be incommensurable, and 
let them be (A +a) and B respectively. Let DE be a line 
divided at C so that 

(4+a):B=DC: CE. 


D Cc E 
a 


Then, if (A +a) placed at # and B placed at D do not 
balance about C, (A +a) is either too great to balance B, or not 
great enough. 

Suppose, if possible, that (A +) is too great to balance B. 
Take from (A +a) a magnitude a smaller than the deduction 
which would make the remainder balance B, but such that the 
remainder A and the magnitude B are commensurable. 


Ἢ A. 13 


194 ARCHIMEDES 


Then, since A, B are commensurable, and 
Air BaD + CEH, 
A and B will not balance [Prop. 6], but D will be depressed. 

But this is impossible, since the deduction ὦ was an 
insufficient deduction from (A +a) to produce equilibrium, so 
that EH was still depressed. 

Therefore (A +a) is not too great to balance B; and 
similarly it may be proved that B is not too great to balance 
(A +a). 

Hence (A+a), B taken together have their centre of 
gravity at Ὁ. 


Proposition 8. 


If AB be a magnitude whose centre of gravity is C, and AD 
a part of it whose centre of gravity 1s F, then the centre of 
gravity of the remaining part will be a point G on FC produced 


such that 
GC : CF=(AD): (DE). 


D B 


For, if the centre of gravity of the remainder (DZ) be not 
G, let it be a point H. Then an absurdity follows at once from 
Props. 6, 7. 


Proposition 9. 


The centre of gravity of any parallelogram lies on the 
straight line joining the middle points of opposite sides. 

Let ABCD be a parallelogram, and let HF join the middle 
points of the opposite sides AD, BC. 

If the centre of gravity does not lie on HF, suppose it to be 
H, and draw HK parallel to AD or BC meeting HF in K. 


ON THE EQUILIBRIUM OF PLANES I. 195 


Then it is possible, by bisecting HD, then bisecting the 
halves, and so on continually, to arrive at a length HL less 


A ΕΞΕΞΕ D 


B F 2 


than KH. Divide both A# and LD into parts each equal 
to HL, and through the points of division draw parallels to AB 
or CD. 


We have then a number of equal and similar parallelograms, 
and, if any one be applied to any other, their centres of gravity 
coincide [Post. 4]. Thus we have an even number of equal 
magnitudes whose centres of gravity lie at equal distances along 
a straight line. Hence the centre of gravity of the whole 
parallelogram will le on the line joining the centres of gravity 
of the two middle parallelograms [Prop. 5, Cor. 2]. 

But this is impossible, for H is outside the middle 
parallelograms. 


Therefore the centre of gravity cannot but lie on HF. 


Proposition 10. 


The centre of gravity of a parallelogram is the point of 
intersection of its diagonals. 

For, by the last proposition, the centre of gravity lies on 
each of the lines which bisect opposite sides. Therefore it 
is at the point of their intersection; and this is also the 
point of intersection of the diagonals. 


Alternative proof. 

Let ABCD be the given parallelogram, and BD a diagonal. 
Then the triangles ABD, CDB are equal and similar, so that 
[ Post. 4], if one be applied to the other, their centres of gravity 
will fall one upon the other. 


13—2 


196 ARCHIMEDES 


Suppose F to be the centre of gravity of the triangle ABD. 
Let G be the middle point of BD. 
Jom FG and produce it to H, so 
that FG = GH. 

If we then apply the triangle 
ABD to the triangle CDB so that 
AD falls on CB and AB on CD,the 8&8 ο 
point F will fall on H. 

But [by Post. 4] # will fall on the centre of gravity of 
CDB. Therefore H is the centre of gravity of CDB. 

Hence, since δ᾽, H are the centres of gravity of the two 
equal triangles, the centre of gravity of the whole parallelogram 
is at the middle point of FH, 1.6. at the middle point of BD, 
which is the intersection of the two diagonals. 


A D 


Proposition 11. 


Tf abe, ABC be two similar triangles, and g, G two points in 
them similarly situated with respect to them respectively, then, if 
g be the centre of gravity of the triangle abc, G must be the centre 
of gravity of the triangle ABC. 


Suppose ab: bec: ca=AB: BC: CA. 


ὃ c B Cc 


The proposition is proved by an obvious reductio ad 
absurdum. For, if G be not the centre of gravity of the 
triangle ABC, suppose H to be its centre of gravity. 

Post. 5 requires that g, H shall be similarly situated with 
respect to the triangles respectively; and this leads at once 
to the absurdity that the angles HAB, GAB are equal. 


ON THE EQUILIBRIUM OF PLANES I. 197 


Proposition 12. 


Given two similar triangles abc, ABC, and d, D the middle 
points of be, BC respectively, then, if the centre of gravity of abc 
lie on ad, that of ABC will lie on AD. 


Let g be the point on ad which is the centre of gravity 
of abe. 
A 


Vega 


b d c B D Cc 


Take G on AD such that 
ad -ag=AD: AG, 
and join gb, gc, GB, GC. 
Then, since the triangles are similar, and bd, BD are the 
halves of bc, BO respectively, 
ab: bd=AB: BD, 
and the angles abd, ABD are equal. 
Therefore the triangles abd, ABD are similar, and 
Zbad = 2 BAD. 
Also ba: ad=BA : AD, 
while, from above, ad: ag=AD: AG. 


Therefore ba: ag=BA: AG, while the angles bag, BAG 
are equal. 


Hence the triangles bag, BAG are similar, and 


Zabg = 2 ABG. 
And, since the angles abd, ABD are equal, it follows that 
Z gbd = 2 GBD. 
In exactly the same manner we prove that 
Mgae— 7G O; 
Zacg=2Z ACG, 


Zgcd—Z2G6CD; 


198 ARCHIMEDES 


Therefore g, G are similarly situated with respect to the 
triangles respectively; whence [Prop. 11] @ is the centre of 
gravity of ABC. 


Proposition 13. 


In any triangle the centre of gravity lies on the straight line 
joining any angle to the middle point of the opposite side. 

Let ABC be a triangle and D the middle point of BC. 
Join AD. Then shall the centre of gravity lie on AD. 

For, if possible, let this not be the case, and let H be the 
centre of gravity. Draw HI parallel to CB meeting AD in J. 

Then, if we bisect DC, then bisect the halves, and so on, 
we shall at length arrive at a length, as DE, less than HI. 


Divide both BD and DC into lengths each equal to DH, and 
through the points of division draw lines each parallel to DA 
meeting BA and AC in points as K, L, M and N, P, Q 
respectively. 


Join MN, LP, KQ, which lines will then be each parallel 
to BC. 


We have now a series of parallelograms as FQ, TP, SN, 
and AD bisects opposite sides in each. Thus the centre 
of gravity of each parallelogram lies on AD [Prop. 9], and 
therefore the centre of gravity of the figure made up of them 
all les on AD. 


ON THE EQUILIBRIUM OF PLANES I. 199 


Let the centre of gravity of all the parallelograms taken 
together be Ὁ. Join OH and produce it; also draw CV 
parallel to DA meeting OH produced in JV. 

Now, if n be the number of parts into which AC is divided, 

AADC : (sum of triangles on AN, VP, ...) 
ea = CANS FN Poe.) 


Ξε "ἃ Τὸ 
Ξε 
eA) = AN. 
Sunilarly 
A ABD : (sum of triangles on AM, ML,...)\=AB: AM. 
And AC ey — Ab: AM. 


It follows that 
A ABC: (sum of all the small As)=CA: AN 
> VO: OG, by parallels. 
Suppose OV produced to X so that 
A ABC: (sum of small As)= XO : OH, 
whence, dividendo, 

(sum of parallelograms) : (sum of small As)= XH: HO. 
Since then the centre of gravity of the triangle ABC is at H, 
and the centre of gravity of the part of it made up of the 
parallelograms is at O, it follows from Prop. 8 that the centre 


of gravity of the remaining portion consisting of all the small 
triangles taken together is at X. 


But this is impossible, since all the triangles are on one side 
of the line through X parallel to AD. 


Therefore the centre of gravity of the triangle cannot but 
lie on AD. 


Alternative proof. 


Suppose, if possible, that H, not lying on AD, is the centre 
of gravity of the triangle ABC. Join AH, BH, CH. Let 
E, F be the middle points of CA, AB respectively, and join 
DE, EF, FD. Let EF meet AD in M. 


200 ARCHIMEDES 


Draw FK, EL parallel to AH meeting BH, CH in K, L 
respectively. Jom KD, HD, LD, KI. Let KL meet DH in 
N, and join MN. 


Since DE is parallel to AB, the triangles ABC, EDC are 
similar. 

And, since CE= FA, and EL is parallel to AH, it follows 
that CL=LH. And CD=DB. Therefore BH is parallel 
to DL. 

Thus in the similar and similarly situated triangles ABC, 
EDC the straight lines AH, BH are respectively parallel to 
EL, DL; and it follows that H, Z are similarly situated with 
respect to the triangles respectively. 

But H is, by hypothesis, the centre of gravity of ABC. 
Therefore Z is the centre of gravity of EDC. [Prop. 11] 

Similarly the point K is the centre of gravity of the 
triangle FBD. 

And the triangles FBD, EDC are equal, so that the centre 
of gravity of both together is at the middle point of KZ, 1.e..at 
the point JV. 

The remainder of the triangle ABC, after the triangles FBD, 
EDC are deducted, is the parallelogram AFDE, and the centre 
of gravity of this parallelogram is at 77, the intersection of its 
diagonals. 

It follows that the centre of gravity of the whole triangle 
ABC must lie on WN; that is, MN must pass through H, which 
is impossible (since JW is parallel to AH). 

Therefore the centre of gravity of the triangle ABC cannot 
but lhe on AD. 


ON THE EQUILIBRIUM OF PLANES I. 201 


Proposition 14. 


It follows at once from the last proposition that the centre 
of gravity of any triangle vs at the intersection of the lines drawn 
from any two angles to the middle points of the opposite sides 
respectively. 


Proposition 15. 


If AD, BC be the two parallel sides of a trapezium ABCD, 
AD being the smaller, and if AD, BC be bisected at KH, F 
respectively, then the centre of gravity of the trapezium ws at a 
point G on EF such that 

GE: GF=(2BC+ AD): 2AD+ BC). 

Produce BA, CD to meet at Ὁ. Then FF produced will 

also pass through O, since AH = HD, and BF = FC. 


Now the centre of gravity of the triangle OAD will lie on 
OL, and that of the triangle OBC will lie on OF. [Prop. 13] 


It follows that the centre of gravity of the remainder, the 
trapezium ABCD, will also lie on OF. [ Prop. 8] 


Join BD, and divide it at LZ, M into three equal parts. 
Through L, M draw PQ, RS parallel to BC meeting BA in 
P, k, FE in W, V, and CD in Q, S respectively. 

Join DF, BE meeting PQ in H and RS in K respectively. 

Now, since BL=1 BD, 

FH =3 FD. 


202 ARCHIMEDES. 


Therefore H is the centre of gravity of the triangle DBC*. 

Similarly, since HK =4 BE, it follows that K is the centre 
of gravity of the triangle ADB. 

Therefore the centre of gravity of the triangles DBC, ADB 
together, 1.6. of the trapezium, les on the line HK. 

But it also lies on OF. 

Therefore, if OF, HK meet in G, G is the centre of gravity 
of the trapezium. 

Hence [Props. 6, 7] 

ADB NABD — KG : Git 


—=VG:GW. 
But A DBC: AABD= BC: AD. 
Therefore BC: AD=VG: GW. 


It follows that 
(2BC + AD): (2AD+ BC)=(2VG+GW) : (2GW + VG) 


= HG: GF. 
Q. E. D. 


* This easy deduction from Prop. 14 is assumed by Archimedes without 
proof. 


ON THE EQUILIBRIUM OF PLANES. 


BOOK II. 


Proposition 1. 


If P,P’ be two parabolic segments and D, E their centres 
of gravity respectively, the centre of gravity of the two segments 
taken together will be at a point C on DE determined by the 
relation 

PP SCH CD*. 

In the same straight line with DH measure HH, EL each 
equal to DC, and Dk equal to DH; whence it follows at once 
that DK =CE, and also that KC = CL. 


* This proposition is really a particular case of Props. 6, 7 of Book I. and 
is therefore hardly necessary. As, however, Book II. relates exclusively to 
parabolic segments, Archimedes’ object was perhaps to emphasize the fact 
that the magnitudes in I. 6, 7 might be parabolic segments as well as 
rectilinear figures. His procedure is to substitute for the segments rect- 
angles of equal area, a substitution which is rendered possible by the results 
obtained in his separate treatise on the Quadrature of the Parabola. 


204 ARCHIMEDES 


Apply a rectangle MN equal in area to the parabolic 
segment P to a base equal to KH, and place the rectangle so 
that KH bisects it, and is parallel to its base. 


Then D is the centre of gravity of MN, since KD= DH, 
Produce the sides of the rectangle which are parallel to KH, 


and complete the rectangle NO whose base is equal to HZ. 
Then F is the centre of gravity of the rectangle VO. 


Now (MN) = NVO)= KA : AL 
OE ΗΝ 
=O CD) 
Sega y 

But (MN) = P. 

Therefore (NO) =P". 


Also, since C is the middle point of AKL, C is the centre 
of gravity of the whole parallelogram made up of the two 
parallelograms (MN), (VO), which are equal to, and have the 
same centres of gravity as, P, P’ respectively. 


Hence C is the centre of gravity of P, P’ takeu together. 


Definition and lemmas preliminary to Proposition 2. 


“Tf in a segment bounded by a straight line and a section 
of a right-angled cone [a parabola] a triangle be inscribed 
having the same base as the segment and equal height, if again 
triangles be inscribed in the remaining segments having the 
same bases as the segments and equal height, and if in the 
remaining segments triangles be inscribed in the same manner, 
let the resulting figure be said to be inscribed in the 
recognised manner (γνωρίμως ἐγγράφεσθαι) in the segment. 


And it is plain 


(1) that the lines jouning the two angles of the figure so inscribed 
which are nearest to the vertex of the segment, and the neat 


ON THE EQUILIBRIUM OF PLANES II. 205 


pairs of angles in order, will be parallel to the base of the 
segment, 

(2) that the said lines will be bisected by the diameter of the 
segment, and 

(3) that they will cut the diameter in the proportions of the 
successive odd numbers, the number one having reference to [the 
length adjacent to] the vertex of the segment. 

And these properties will have to be proved in their proper 
places (ἐν ταῖς τάξεσιν). 

[The last words indicate an intention to give these pro- 
positions in their proper connexion with systematic proofs; but 
the intention does not appear to have been carried out, or at 
least we know of no lost work of Archimedes in which they 
could have appeared. The results can however be easily 
derived from propositions given in the Quadrature of the 
Parabola as follows. 

(1) Let BRQPApgqrb be a figure inscribed ‘in the recog- 
nised manner’ in the parabolic segment BAb of which Bb is 
the base, A the vertex and AO the diameter. 


Bisect each of the lines BQ, BA, QA, Aq, Ab, gb, and 
through the middle points draw lines parallel to AO meeting 
Bb in G, F, E, e, f, g respectively. 


206 ARCHIMEDES 


These lines will then pass through the vertices R, Q, P, 
p,q, 7 of the respective parabolic segments [Quadrature of the 
Parabola, Prop. 18], 1.6. through the angular points of the 
inscribed figure (since the triangles and segments are of equal 


height). 


Also BG=GF=FE=EO0, and Oe=ef=fg=g9b. But 
BO = Ob, and therefore all the parts into which Bb is divided 


are equal. 
If now AB, RG meet in ἢν, and Ab, rg in l, we have 
BG: GL=B0O : OA, by parallels, 
=b0: 0A 


=bg: gl, 
whence GL = gl. 


Again [zbid., Prop. 4] 
Gi: Lh= BO: OG 
=b0: Og 


ΞΞ peline 
and, since GL=gl, LR=lIr. 


Therefore GR, gr are equal as well as parallel. 
Hence GRrg is a parallelogram, and Fr is parallel to Bo. 


Similarly it may be shown that Pp, Qq are each parallel 
to Bb. 


(2) Since RGgr is a parallelogram, and RG, rg are 
parallel to AO, while GO= Og, it follows that Rr is bisected 
by AO. 

And similarly for Pp, Qq. 

(3) Lastly, if V, W, X be the points of bisection of Pp, 
Qq, Rr, 

AVZAW : AX -AO=PV- 2 OW hae ΡΣ 
= ΟΞ 10; 
whence AVEVW : WX 2 ΧΟ ΞΘ Ὁ ia 


ON THE EQUILIBRIUM OF PLANES II. 207 


Proposition 2. 


If a figure be ‘inscribed in the recognised manner’ in a 
parabolic segment, the centre of gravity of the figure so inscribed 
will lie on the diameter of the segment. 

For, in the figure of the foregoing lemmas, the centre of 
gravity of the trapezium BRrb must lie on XO, that of the 
trapezium RQgr on WX, and so on, while the centre of gravity 
of the triangle PAp lies on AV. 


Hence the centre of gravity of the whole figure lies on 40. 


Proposition S. 


If BAB’, bab’ be two similar parabolic segments whose 
diameters are AO, ao respectively, and if a figure be inscribed 
in each segment ‘in the recognised manner, the number of sides 
in each figure being equal, the centres of gravity of the inscribed 
figures will divide AO, ao in the same ratio. 


[Archimedes enunciates this proposition as true of similar 
segments, but it is equally true of segments which are not 
similar, as the course of the proof will show.] 

Suppose BRQPAP’Q'R’B’, brqpap’qr'b’ to be the two 
figures inscribed ‘in the recognised manner. Join PP’, QQ’, 
RR’ meeting AO in ἢ, M, N, and pp’, qq’, rr’ meeting ao 
in 1, m, n. 

Then [Lemma (3)] 

Al, = Divi = NO 
0 5:7 
=al:lm: mn: no, 
so that AO, ao are divided in the same proportion. 
Also, by reversing the proof of Lemma (3), we see that 
PP pp OO gen or = BB’ - bb. 

Since then RR’: BB'=rr’: bb’, and these ratios respec- 

tively determine the proportion in which NO, no are divided 


208 ARCHIMEDES 


by the centres of gravity of the trapezia BRR’B’, brr’b’ [τ 15], 
it follows that the centres of gravity of the trapezia divide VO, 
no in the same ratio. 


Similarly the centres of gravity of the trapezia RQQ’R’, 
rqq” divide MN, mn in the same ratio respectively, and so on. 

Lastly, the centres of gravity of the triangles PAP’, pap’ 
divide AZ, al respectively in the same ratio. 

Moreover the corresponding trapezia and triangles are, each 
to each, in the same proportion (since their sides and heights 
are respectively proportional), while AO, ao are divided in 
the same proportion. 

Therefore the centres of gravity of the complete inscribed 
figures divide AO, ao in the same proportion. 


Proposition 4. 


The centre of gravity of any parabolic segment cut off by a 
straight line lies on the diameter of the segment. 

Let BAB’ be a parabolic segment, A its vertex and AO its 
diameter. 

Then, if the centre of gravity of the segment does not lie on 
AO, suppose it to be, if possible, the point Κ΄. Draw FE 
parallel to AO meeting BB’ in ἢ. 


ON THE EQUILIBRIUM OF PLANES II. 209 


Inscribe in the segment the triangle ABB’ having the same 
vertex and height as the segment, and take an area S such 
that 

AABB’: S=BE: EO. 


L 


K B 


B’ 


We can then inscribe in the segment ‘in the recognised — 
manner’ a figure such that the segments of the parabola left 
over are together less than S. [For Prop. 20 of the Quadrature 
of the Parabola proves that, if in any segment the triangle with 
the same base and height be inscribed, the triangle is greater 
than half the segment; whence it appears that, each time that 
we increase the number of the sides of the figure inscribed ‘in 
the recognised manner, we take away more than half of the 
remaining segments. | 


Let the inscribed figure be drawn accordingly ; its centre 
of gravity then lies on AO [Prop. 2]. Let it be the point H. 


Join HF and produce it to meet in K the line through B 
parallel to AO. 


Then we have 
(inscribed figure) : (remainder of segmt.) > A ABB’: S 
>BE: EO 
So) i hae Wal: 


Suppose Z taken on HK produced so that the former ratio is 
equal to the ratio LF : FH. 


H. A. 14 


210 ARCHIMEDES 


Then, since H is the centre of gravity of the inscribed 
figure, and F# that of the segment, Z must be the centre 
of gravity of all the segments taken together which form the 
remainder of the original segment. [I. 8] 


But this is impossible, since all these segments lie on one 


side of the line drawn through ZL parallel to AO [Cf. Post. 7]. 


Hence the centre of gravity of the segment cannot but lie 
on AO. 


Proposition 5. 


If in a parabolic segment a figure be inscribed ‘in the 
recognised manner, the centre of gravity of the segment 1s nearer 
to the verter of the segment than the centre of gravity of the 
inscribed figure 18. 


Let BAB’ be the given segment, and AO its diameter. 
First, let ABB’ be the triangle in- 
scribed ‘in the recognised manner.’ 


Divide AO in F so that AF=2F0; 
F is then the centre of gravity of the 
triangle ABB’. 

Bisect AB, AB’ in D, D' respec- 
tively, and jom DD’ meeting AO in £. 
Draw DQ, D’Q’ parallel to OA to meet 
the curve. QD, Q'D’ will then be the 
diameters of the segments whose bases 
are AB, AB’, and the centres of gravity 
of those segments will lie respectively 
on QD, Q’D’ [Prop. 4]. Let them be ἢ, H’, and join HH’ 
meeting AO in ζ΄. 


B 


Now QD, Q’D’ are equal*, and therefore the segments of 
which they are the diameters are equal [On Conoids and 
Spheroids, Prop. 3]. 


* This may either be inferred from Lemma (1) above (since QQ’, DD’ are 
both parallel to BB’), or from Prop. 19 of the Quadrature of the Parabola, which 
applies equally to Q or Q’. 


ON THE EQUILIBRIUM OF PLANES II. rAd 


Also, since QD, Q’D’ are parallel*, and DE = ED’, K is the 
middle point of HH’. 

Hence the centre of gravity of the equal segments AQB, 
AQ’B’ taken together is K, where K les between # and A. 
And the centre of gravity of the triangle ABB’ is F. 


It follows that the centre of gravity of the whole segment 
BAB’ lies between K and F, and is therefore nearer to the 
vertex A than F' is. 

Secondly, take the five-sided figure BQAQ’B’ inscribed ‘in 
the recognised manner, QD, Q’D’ being, as before, the diameters 


of the segments AQB, AQ’B’. 


Then, by the first part of this proposition, the centre of 
gravity of the segment AQB (lying of course on QD) is nearer 
to Q than the centre of gravity of the triangle AQB is. Let 
the centre of gravity of the segment be H, and that of the 
triangle J. 


Similarly let H’ be the centre of gravity of the segment 
AQ’B’, and I’ that of the triangle AQ’B’. 

It follows that the centre of gravity 
of the two segments AQB, AQ’ B’ taken 
together is Δ΄, the middle point of HH’, 
and that of the two triangles AQB, AQ’B’ 
is L, the middle point of JI’. 


If now the centre of gravity of the 
triangle ABB’ be F, the centre of gravity 
of the whole segment BAB’ (i.e. that of 
the triangle ABB’ and the two segments 
AQB, AQ'B’ taken together) is a point 
G on KF determined by the proportion 


(sum of segments AQB, AQ’B’): AABB’=FG: GK. [I. 6,7] 


* There is clearly some interpolation in the text here, which has the words 
kal ἐπεὶ παραλληλόγραμμόν ἐστι τὸ OZHI. It is not yet proved that H’D’DH is 
a parallelogram ; this can only be inferred from the fact that H, H’ divide QD, 
Q'D’ respectively in the same ratio. But this latter property does not appear 
till Prop. 7, and is then only enunciated of similar segments. The interpolation 
must have been made before Eutocius’ time, because he has a note on the 
phrase, and explains it by gravely assuming that H, H’ divide QD, Q’D’ respec- 
tively in the same ratio. 


14—2 


212 ARCHIMEDES 

And the centre of gravity of the inscribed figure BQAQ’B’ 
is a point F’ on LF determined by the proportion 

(AAQB+AAQ'B) :-AABD =F’? FL. “ieee 

[Hence EG Gk Pr: FL, 
or Gi Get Die 
and, componendo, FK:FG<FL: FF’, while FK> FL.) 
Therefore FG > FF’, or G lies nearer than F” to the vertex A. 


Using this last result, and proceeding in the same way, 
we can prove the proposition for any figure inscribed ‘in the 
recognised manner.’ 


Proposition 6. 


Given a segment of a parabola cut off by a straight line, τέ is 
possible to inscribe in it ‘in the recognised manner’ a figure such 
that the distance between the centres of gravity of the segment and 
of the inscribed figure is less than any assigned length. 


Let BAB’ be the segment, AO its diameter, @ its centre 
of gravity, and ABB’ the triangle inscribed ‘in the recognised 
manner.’ 


Let D be the assigned length and S an area such that 
AG: D=AABB > δ. 


In the segment inscribe ‘in the recognised manner’ a figure 
such that the sum of the segments left over is less than S. 
Let F be the centre of gravity of the inscribed figure. 


We shall prove that FG < D. 
For, if not, FG must be either equal to, or greater than, D. 
And clearly 
(inscribed fig.) : (sum of remaining segmts.) 
ABB aS 
SGD 
> AG : FG, by hypothesis (since FG 4 D). 


ON THE EQUILIBRIUM OF PLANES II. 213 


Let the first ratio be equal to the ratio KG : FG (where K 
lies on GA produced); and it follows that K is the centre of 
gravity of the small segments taken together. [I. 8] 


B 


Oo 


But this is impossible, since the segments are all on the 
same side of a line drawn through K parallel to BB’. 


Hence FG cannot but be less than D. 


Proposition 7. 


If there be two similar parabolic segments, their centres of 
gravity divide their diameters in the same ratio. 


[This proposition, though enunciated of similar segments 
only, like Prop. 3 on which it depends, is equally true of 
any segments. This fact did not escape Archimedes, who 
uses the proposition in its more general form for the proof of 
Prop. 8 immediately following. ] 


Let BAB’, bab’ be the two similar segments, AO, ao their 
diameters, and G, g their centres of gravity respectively. 


Then, if G, g do not divide AO, ao respectively in the same 
ratio, suppose H to be such a point on AO that 


AH : HO=ag: go; 


214 ARCHIMEDES 


and inscribe in the segment BAB’ ‘in the recognised manner’ 
a figure such that, if F be its centre of gravity, 


GF <GH. [Prop. 6] 


Inscribe in the segment bab’ ‘in the recognised manner’ a 
similar figure; then, if f be the centre of gravity of this figure, 


ag <af. [ Prop. 5] 
And, by Prop. 3, af: fo=AF: FO. 
But er Ah O <= AHO 


< ag : go, by hypothesis. 
Therefore af : fo<ag : go; which is impossible. 
It follows that G, g cannot but divide AO, ao in the same 
ratio. 
Proposition 8. 


If AO be the diameter of a parabolic segment, and G its 

centre of gravity, then 
AG=3 GO. 

Let the segment be BAB’. Inscribe the triangle ABB’ ‘in 
the recognised manner,’ and let Κ᾽ be its centre of gravity. 

Bisect AB, AB’ in ἢ, D’,and draw DQ, D’Q’ parallel to OA 
to meet the curve, so that QD, Q’D’ are the diameters of the 
segments AQB, AQ’B’ respectively. 

Let H, H’ be the centres of gravity of the segments AQB, 
AQ’B’ respectively. Join QQ’, HH’ meeting AO in V, Καὶ 
respectively. 


ON THE EQUILIBRIUM OF PLANES II. 215 
K is then the centre of gravity of the two segments AQB, 
AQ’ B’ taken together. 
Now “0 αὐ ΞΌΗΣ 7D) 
[Prop. 7] 
whence AGOGO DED: 


But AO=4QD [as is easily proved 
by means of Lemma (3), p. 206]. 


Therefore OG=4HD 
and, by subtraction, AG=4Q#H. 
Also, by Lemma (2), QQ’ is paral- 


lel to BB’ and therefore to DD’. It 
follows from Prop. 7 that HH" is also parallel to QQ’ or DD’, 


and hence OF= VK, 
Therefore AG=4VK, 
and AV+KG=3VK. 
Measuring VL along VK so that VZ=1AV, we have 
LG BUD SA a ere ener OP (1) 
Again AO=4AV [Lemma (3)] 
=3AL, since AV=3VEL, 
whence A ΞΞ ΤΡ OOM assis cine ceeraine « (2). 
Now, by I. 6, 7, 


A ABB’ : (sum of segmts. AQB, AQ’ B’)= Κα: GF, 
and A ABB’ =3 (sum of segments AQB, AQ’B’) 


[since the segment ABB’ is equal to 4 A ABB’ (Quadrature of 
the Parabola, Props. 17, 24)]. 


Hence KG=3GF. 
But KG =3LK, from (1) above. 
Therefore LF=LK+KG+GPr 


216 ARCHIMEDES 


And, from (2), 
LF=(AO-—AL-— OF)=1A0= OF. 


Therefore OF =5GF, 
and OG =6GF. 
But AQ=sOF = 15GF. 
Therefore, by subtraction, 
AG=9GF 
=O, 


Proposition 9 (Lemma). 
If a, ὃ, c, d be four lines in continued proportion and in 
descending order of magnitude, and uf 
d:(a—d)=2:3(a—-c), 
and (2a+4b+6c+ 3d) : (5a+10b+4+ 10c + 5d)=y : (a—c), 
at is required to prove that 


e+y=2a. 
[The following is the proof given by Archimedes, with 
the only difference that it is set out in A 


algebraical instead of geometrical notation. 
This is done in the particular case simply in 
order to make the proof easier to follow. Τ' 
Archimedes exhibits his lines in the figure 
reproduced in the margin, but, now that it is 
possible to use algebraical notation, there is a 
no advantage in using the figure and the more 
cumbrous notation which only obscures the course 
of the proof. The relation between Archimedes’ 
figure and the letters used below is as follows; B 
AB=a, TB=6b, AB=c, EB=d, ZH =a, HO=y, AO =z.] 


We have > Sa SSE ee τε πα eens (1), 


mo 


whence τ-- 


and therefore i - -- τ 


2(a+b)_at+b_a+b δ a-c b-—c_a—¢ 


Bow 2c Ca Uh 6 (Ob Seve ame 


ON THE EQUILIBRIUM OF PLANES II. ὉΠ 7 


And, in like manner, 
ἜΠΟΣ ἘΠ eee 
d c ‘ad δα 
It follows from the last two relations that 
— = s ἊΣ CC ane (3). 
Suppose z to be so taken that 


2α -ἰ 40 -- 4. - 24 a-—c 
το πώ eee e eee e er eeeee (4), 
so that z<(c—d). 
a-c+z_ 2a+4b+6c+3d 
a—c 2(a+d)+4(b+c) 
And, by hypothesis, 
a-c_5(a+d)+10(b+¢) 
y 2a+4b+6c+3d ᾿ 
a—c+z_5(a+d)+10(b+c)_ 5 3 
y EEE abc) Bn 
Again, dividing (3) by (4) crosswise, we obtain 


Therefore 


so that 


as 2a+3b+¢ 
c—d 2(a+d)+4(b+c)’ 
πα-- 2 b+ 36- 2d 
| Gian ices oO 
whence a τ τὴ (0). 
But, by (2), 
c-d_a—b_3(b-c)_2(¢-d) 
d oe | | od 5.” 
c—d_ (a—b)+3(b—c)+2(c—d) 
so that ee _aege ΤΗΝ ον (7. 


Combining (6) and (7), we have 
e-d—z_(a—b)+3(b—c)+2(c—d) 
d 2 (α -- d)+4(b+c) Ἶ 
6-ιΖ 3a + 6b + 30 
h = 
W cine d 2(at+d)+4(b +0) eieotete vieleielelevelorate 
And, since [by (1)] 
c-~d_b-c_ a—b 
c+d b+e a+b’ 


218 ARCHIMEDES 


c—d c+d 
we have — 

a—c b+c+a+b’ 
a-—d_a+2b+2c+d_ 2(a+d)+4(b+c) 
prone —-c a@+2b+¢  2(a+c)+4b 

Thus a—d  2(a+d)+4(b+c) 


g(a—0) 2 (2(atc) +H} ” 
and therefore, by hypothesis, 

αὐ 2(a+d)+4(b+c) 

a  #{2(a+c)+ 4d} 
c—z 3a+6b+3c | 

d  2(a+d)+4(b+c)’ 
and it follows, ex aequalt, that 


But, by (8), 


C= 2 θη (ΟΕ. δ 5. ὦ 
a 8(2(α 0}: 40) 3°2 2° 
A P eh ae 
nd, by (5), i δ᾽ 
Therefore = δ εἰ 
a @+Y 
or a+y Ξε ξα. 


Proposition 10. 

If PP’B’B be the portion of a parabola intercepted between 
two parallel chords PP’, BB’ bisected respectively in N, O by 
the diameter ANO (N being nearer than O to A, the vertex 
of the segments), and if NO be divided into five equal parts of 
which LM is the middle one (L being nearer than M to N), then, 
of G be a point on LM such that 

LG: GM= BO’ .(2PN + BO): PN’ .(2B0+ PN), 
G will be the centre of gravity of the area PP’ Β' B. 
Take a line ao equal to AO, and an on it equal to AN. Let 
p, ¢ be points on the line ao such that 
ὦ ἀξ GW saticees scene (1), 
ο΄: ON = CG QD sass eens ata (2), 
[whence ao : ag = aq : an=an: ap, or ao, aq, an, ap are lines in 
continued proportion and in descending order of magnitude]. 
Measure along GA a length GF such that 
Op Gp = ΟΠ (3). 


ON THE EQUILIBRIUM OF PLANES II. 219 


Then, since PN, BO are ordinates to ANO, 
ΒΩ ΝΞΟ AN 
ΞΕ Ὁ : (1) 
Ξε 0.200.) Δ (L), 
so that ποτ΄ Π΄Π-----. (4), 
and BO’; PN = ao: aq" 
= (ao : ag).(aq : an). (an : ap) 
ΞΞ Ὁ : ap 


Β' 
Thus (segment BAB’) : (segment PAP’) 
=) A Jos Βα. aera aah 
ΞΡ" 


= do: ap, 
whence 
(area PP’B’B) : (segment PAP’) =op : ap 
=OL: GF, by (8), 
ΞΞ AGE 2 sons ros (6). 
Now BO’ .(2PN + BO): BO’=(2PN + BO): BO 
=(2aq+ ao) : ao, by (4), 
BO*: PN*=ao : ap, by (5), 
and ΡΝ PN @B0-aPN)—PN. (280+ PN) 
Ξε αᾳ : (2a0+aq), by (4), 
= ap: (2an+ap), by (2). 


220 ARCHIMEDES. 


Hence, ex aequalz, 
BO’ .(2PN + BO): PN’. (2B0 + PN) =(2aq + ao) : (2an + ap), 
so that, by hypothesis, 
LG : GM =(2aq + ao) : (2an + ap). 
Componendo, and multiplying the antecedents by 5, 
ΟΝ : GM= {5 (ao+ap)+10 (aq +an)} : (2an+ ap). 
But ON: OM=5:2 
= {5(ao+ap)+10(aq+an)} : {2(ao+ap)+4(aq+an)}. 

It follows that 
ON : OG ={5 (ao + ap) +10 (aq +an)} : (2a0 + 4aq + 6an + 3ap). 
Therefore 
(2a0 + 4aq + Gan + 8ap) : {5 (ao+ap)+ 10 (aq +an)} = OG : ON 


= 0G on 
And ap : (ao — ap) =ap : op 
= GF : OL, by hypothesis, 
= GF : on, 


while ao, aq, an, ap are in continued proportion. 
Therefore, by Prop. 9, 
GF + OG = OF = 2a0=20A. 
Thus F is the centre of gravity of the segment BAB’. [Prop. 8] 


Let H be the centre of gravity of the segment PAP’, so 
that AH =2AN. 


And, since Af =2A0, 
we have, by subtraction, HF =20N. 
But, by (6) above, 
(area PP’B’B) : (segment PAP’) =80N: GF 
ΞΞ δ. THE: 


Thus, since 1΄, H are the centres of gravity of the segments 
BAB’, PAP’ respectively, it follows [by I. 6, 7] that G is the 
centre of gravity of the area PP’B’B. 


THE SAND-RECKONER. 


“THERE are some, king Gelon, who think that the number 
of the sand is infinite in multitude; and I mean by the sand 
not only that which exists about Syracuse and the rest of Sicily 
but also that which is found in every region whether inhabited 
or uninhabited. Again there are some who, without regarding 
it as infinite, yet think that no number has been named which 
is great enough to exceed its multitude. And it is clear that 
they who hold this view, if they imagined a mass made up of 
sand in other respects as large as the mass of the earth, in- 
cluding in it all the seas and the hollows of the earth filled up 
to a height equal to that of the highest of the mountains, 
would be many times further still from recognising that any 
number could be expressed which exceeded the multitude of 
the sand so taken. But I will try to show you by means of 
geometrical proofs, which you will be able to follow, that, of the 
numbers named by me and given in the work which I sent to 
Zeuxippus, some exceed not only the number of the mass of 
sand equal in magnitude to the earth filled up in the way 
described, but also that of a mass equal in magnitude to the 
universe. Now you are aware that ‘universe’ is the name 
given by most astronomers to the sphere whose centre is the 
centre of the earth and whose radius is equal to the straight 
line between the centre of the sun and the centre of the earth. 
This is the common account (τὰ γραφόμενα), as you have heard 
from astronomers. But Aristarchus of Samos brought out a 


222 ARCHIMEDES 


book consisting of some hypotheses, in which the premisses lead 
to the result that the universe is many times greater than that 
now so called. His hypotheses are that the fixed stars and the 
sun remain unmoved, that the earth revolves about the sun in 
the circumference of a circle, the sun lying in the middle of the 
orbit, and that the sphere of the fixed stars, situated about. the 
same centre as the sun, is so great that the circle in which he 
supposes the earth to revolve bears such a proportion to the 
distance of the fixed stars as the centre of the sphere bears to 
its surface. Now it is easy to see that this is impossible; for, 
since the centre of the sphere has no magnitude, we cannot 
conceive it to bear any ratio whatever to the surface of the 
sphere. We must however take Aristarchus to mean this: 
since we conceive the earth to be, as it were, the centre of 
the universe, the ratio which the earth bears to what we 
describe as the ‘universe’ is the same as the ratio which the 
sphere containing the circle in which he supposes the earth to 
revolve bears to the sphere of the fixed stars. For he adapts 
the proofs of his results to a hypothesis of this kind, and in 
particular he appears to suppose the magnitude of the sphere 
in which he represents the earth as moving to be equal to what 
we call the ‘ universe.’ 

I say then that, even if a sphere were made up of the sand, 
as great as Aristarchus supposes the sphere of the fixed stars 
to be, I shall still prove that, of the numbers named in the 
Principles*, some exceed in multitude the number of the 
sand which is equal in magnitude to the sphere referred to, 
provided that the following assumptions be made. 

1. The perimeter of the earth is about 3,000,000 stadia and 
not greater. 

It is true that some have tried, as you are of course aware, 
to prove that the said perimeter is about 300,000 stadia. But 
I go further and, putting the magnitude of the earth at ten 
times the size that my predecessors thought it, I suppose its 
perimeter to be about 3,000,000 stadia and not greater. 

*’Apyal was apparently the title of the work sent to Zeuxippus. Cf. the 


note attached to the enumeration of lost works of Archimedes in the Introduction, 
Chapter II., ad fin. 


THE SAND-RECKONER. 223 


2. The diameter of the earth is greater than the diameter of 
the moon, and the diameter of the sun is greater than the diameter 
of the earth. 


In this assumption I follow most of the earlier astronomers. 


3. The diameter of the sun is about 30 tumes the diameter of 
the moon and not greater. 


It is true that, of the earlier astronomers, Eudoxus declared 
it to be about nine times as great, and Pheidias my father* 
twelve times, while Aristarchus tried to prove that the diameter 
of the sun is greater than 18 times but less than 20 times the 
diameter of the moon. But I go even further than Aristarchus, 
in order that the truth of my proposition may be established 
beyond dispute, and I suppose the diameter of the sun to be 
about 30 times that of the moon and not greater. 


4. The diameter of the sun ws greater than the side of the 
chiliagon inscribed in the greatest circle in the (sphere of the) 
unwwerse. 


I make this assumption+ because Aristarchus discovered 
that the sun appeared to be about 74,th part of the circle of 
the zodiac, and I myself tried, by a method which I will now 
describe, to find experimentally (ὀργανικῶς) the angle sub- 
tended by the sun and having its vertex at the eye (τὰν γωνίαν, 


εἰς Av ὁ ἅλιος ἐναρμόζει τὰν κορυφὰν ἔχουσαν ποτὶ τᾷ ὄψει). 
ἐς PE P x ν π' C : 


[Up to this point the treatise has been literally translated 
because of the historical interest attaching to the ipsissima 
verba of Archimedes on such a subject. The rest of the work 
can now be more freely reproduced, and, before proceeding to 
the mathematical contents of it, it is only necessary to remark 
that Archimedes next describes how he arrived at a higher and 
a lower limit for the angle subtended by the sun. This he did 


* ro0 ἁμοῦ πατρὸς is the correction of Blass for τοῦ ᾿Ακούπατρος (Jahrb. f. 


Philol. cxxvu1. 1883). 

+ This is not, strictly speaking, an assumption ; it is a proposition proved 
later (pp. 224—6) by means of the result of an experiment about to be 
described. 


224 ARCHIMEDES 


by taking a long rod or ruler (κανών), fastening on the end of it 
a small cylinder or disc, pointing the rod in the direction of the 
sun just after its rising (so that it was possible to look directly 
at it), then putting the cylinder at such a distance that it just 
concealed, and just failed to conceal, the sun, and lastly measur- 
ing the angles subtended by the cylinder. He explains also the 
correction which he thought it necessary to make because “ the 
eye does not see from one point but from a certain area” (ἐπεὶ 
ai ὄψιες οὐκ ad’ ἑνὸς σαμείου βλέποντι, ἀλλὰ ἀπό τινος 
μεγέθεος).] 

The result of the experiment was to show that the angle 
subtended by the diameter of the sun was less than ;4,th part, 
and greater than z4,th part, of a right angle. 


To prove that (on this assumption) the diameter of the sun 
as greater than the side of a chiliagon, or figure with 1000 equal 
sides, inscribed in a great circle of the “ universe.’ 


Suppose the plane of the paper to be the plane passing 
through the centre of the sun, the centre of the earth and the 
eye, at the time when the sun has just risen above the horizon. 
Let the plane cut the earth in the circle HHL and the sun 
in the circle FXG, the centres of the earth and sun being C, O 
respectively, and # being the position of the eye. 


Further, let the plane cut the sphere of the ‘universe’ (i.e. 
the sphere whose centre is C and radius CQ) in the great 
circle AOB. 


Draw from / two tangents to the circle FKG touching it 
at P, Q, and from C draw two other tangents to the same circle 
touching it in F, G respectively. 


Let CO meet the sections of the earth and sun in H, K 
respectively ; and let CF’, CG produced meet the great circle 
AOB in A, B. 


Join HO, OF, OG, OP, OQ, AB, and let AB meet CO in MW. 
Now CO > EO, since the sun is just above the horizon. 
Therefore PHOS ACG. 


THE SAND-RECKONER. 225 


And 2 PEQ> εὐ where 1 represents a right angle. 


but «τς 


Thus ZLECG < 7h, a fortior, 


and the chord AB subtends an are of the great circle which is 
less than ;1,th of the circumference of that circle, ie. 


AB <(side of 656-sided polygon inscribed in the circle). 


Now the perimeter of any polygon inscribed in the great 
circle is less than 4400. [Cf Measurement of a circle, Prop. 3.] 


Therefore ΑΒ: CO < il: 1148. 
and, a fortiors, PAUSE TAO (es cticidats sua sctuisndeiaite (a). 


Again, since CA =CO, and AWM is perpendicular to CO, 
while OF is perpendicular to CA, 


AM = OF. 
Therefore AB=2AM = (diameter of sun). 
Thus (diameter of sun) < 73,00, by (a), 
and, a fortiori, 
(diameter of earth) < ;1,00. [Assumption 2] 


H. A. 15 


226 ARCHIMEDES 


Hence CH + OK < 73,00, 
so that HK > 23,00, 
or CO: HK <100: 99. 
And CO > CF, 
while HK < EQ. 
Therefore CPs 5) Φ 100 2 99:..zc0c5- eee (8). 


Now in the right-angled triangles CFO, EQO, of the sides 
about the right angles, 


OF = OQ, but EQ < CF (since LO < CO). 
Therefore ZOEQ: ZOCF>CO: EO, 
but <OF -EO* 
Doubling the angles, 
ZPEQ:>ZACB < CF : EQ 
< 100 : 99, by (8) above. 


But ZPEQ > zA,R, by hypothesis. 
Therefore LACB>323,R 
> ohh. 


It follows that the arc AB is greater than ,1,th of the circum- 
ference of the great circle AOB. 


Hence, a fortiori, 
AB >(side of chiliagon inscribed in great circle), 


and AB is equal to the diameter of the sun, as proved above. 


The following results can now be proved : 
(diameter of ‘universe’) < 10,000 (diameter of earth), 
and (diameter of ‘ universe’) < 10,000,000,000 stadia. 


* The proposition here assumed is of course equivalent to the trigonometrical 
formula which states that, if a, 8 are the circular measures of two angles, each 
less than a right angle, of which a is the greater, then 


tana .α sina 
tang” B” sing’ 


THE SAND-RECKONER. 997 


(1) Suppose, for brevity, that d, represents the diameter 
of the ‘universe,’ d, that of the sun, d, that of the earth, and d,, 
that of the moon. 


By hypothesis, ds > 30d m, [Assumption 3] 
and ee idin: [Assumption 2] 
therefore ad, <= 308,. 


Now, by the last proposition, 
ds > (side of chiliagon inscribed in great circle), 
so that (perimeter of chiliagon) < 1000d, 
< 30,000d,. 

But the perimeter of any regular polygon with more sides 
than 6 inscribed in a circle is greater than that of the inscribed 
regular hexagon, and therefore greater than three times the 
diameter. Hence 

(perimeter of chiliagon) > 3d,. 

It follows that dy, < 10,000d,. 

(2) (Perimeter of earth) + 3,000,000 stadia. 


[Assumption 1] 
and (perimeter of earth) > 3d,. 


Therefore ας < 1,000,000 stadia, 
whence dy, < 10,000,000,000 stadia. 


Assumption 5. 

Suppose a quantity of sand taken not greater than a poppy- 
seed, and suppose that it contains not more than 10,000 grains. 

Next suppose the diameter of the poppy-seed to be not less 
than {4th of a finger-breadth. 

Orders and periods of numbers. 

I. We have traditional names for numbers up to a 
myriad (10,000); we can therefore express numbers up to a 
myriad myriads (100,000,000). Let these numbers be called 
numbers of the first order. 

Suppose the 100,000,000 to be the unit of the second order, 
and let the second order consist of the numbers from that unit 
up to (100,000,000)”. 

15—2 


228 ARCHIMEDES 


Let this again be the unit of the third order of numbers 
ending with (100,000,000)’; and so on, until we reach the 
100,000,000th order of numbers ending with (100,000,000), 
which we will call P. 


II. Suppose the numbers from 1 to P just described to 
form the first period. 


Let P be the unit of the first order of the second period, and 
let this consist of the numbers from P up to 100,000,000 P. 


Let the last number be the unit of the second order of the 
second period, and let this end with (100,000,000)? P. 


We can go on in this way till we reach the 100,000,000th order 
of the second period ending with (100,000,000)! P, or P?. 


III. Taking P’ as the unit of the first order of the third 
period, we proceed in the same way till we reach the 
100,000,000th order of the third period ending with ἃ 


IV. Taking P* as the unit of the first order of the fourth 
period, we continue the same process until we arrive at the 
100,000,000th order of the 100,000,000th period ending with 
P2,000,000, This last number is expressed by Archimedes as “a 
myriad-myriad units of the myriad-myriad-th order of the 
myriad-myriad-th period (ai μυριακισμυριοστᾶς περιόδου μυρια- 
κισμυριοστῶν ἀριθμῶν μυρίαι μυριάδες), Which is easily seen 
to be 100,000,000 times the product of (100,000,000) and 


P22,999,999 1.e. P 100,000,000. 


[The scheme of numbers thus described can be exhibited 
more clearly by means of indices as follows. 


FIRST PERIOD. 
First order. Numbers from 1 to 10% 
Second order. ve jun) 108 tonto 
Third order. 3 2 10" tote 


(108)th order. ᾿ ». 10. ει aaa 


THE SAND-RECKONER. 229 


SECOND PERIOD. 
First order. Numbers from P.1 to P.108 
Second order. u SESS OE Gore OH, 


(10°)th order. A Spee TOs Ce to 
Ὁ. Τρ (oreP:): 


(108)rH PERIOD. 
First order. τ πο συ to Pe, 70: 
Second order. i Sree 10 te P10”. 


(10°)th order. - es Eas Ue Ὁ τ} 0 
OP! (δ. ἀραὶ 

The prodigious extent of this scheme will be appreciated 
when it is considered that the last number in the first period 
would be represented now by 1 followed by 800,000,000 ciphers, 
while the last number of the (10*)th pertod would require 
100,000,000 times as many ciphers, i.e. 80,000 million millions 
of ciphers. ] 

Octads. 

Consider the series of terms in continued proportion of 
which the first is 1 and the second 10 [ie. the geometrical 
progression 1, 10', 10%, 10°, ...]. The first octad of these terms 
[v.e. 1, 10', 10?,... 107] fall accordingly under the first order 
of the first period above described, the second octad [1.6. 
10°, 10°,... 1015] under the second order of the first period, the 
first term of the octad being the unit of the corresponding 
order in each case. Similarly for the third octad, and so on. 
We can, in the same way, place any number of octads. 

Theorem. 


If there be any number of terms of a series in continued 
proportion, say Ay Asi ΤΠ Amir—,--- Of which 
A,=1, A,=10 [so that the series forms the geometrical pro- 
gression LEO) 107 tn) =e LO LOT... | and af any 
two terms as Am, An be taken and multiplied, the product 


230 ARCHIMEDES 


Am.A, will be a term in the same series and will be as many 
terms distant from Ay as A, is distant from A,; also τέ will be 
distant from A, by a number of terms less by one than the sum 
of the numbers of terms by which A, and A, respectively are 
distant from Ay. 

Take the term which is distant from A, by the same 
number of terms as A,, is distant from A,. This number of 
terms is m (the first and last being both counted). Thus the 
term to be taken is m terms distant from A,, and is therefore 
the term Amin. 


We have therefore to prove that 
isle . Fal = Ar ΥΞΤ: 
Now terms equally distant from other terms in the con- 
tinued proportion are proportional. 


An Ads Amin— 
Thus cae ie 
But PA nA. eA, j Since Agee 
Therefore ΙΕ ΞΡ ate Apna dsetaeds oa eee (1). 


The second result is now obvious, since A, is m terms 
distant from A,, A, is n terms distant from A,, and Aj» in 1S 
(m+n—1) terms distant from A). 


Application to the number of the sand. 
By Assumption 5 [p. 227], 
(diam. of poppy-seed) ¢ 4, (finger-breadth) ; 


and, since spheres are to one another in the triplicate ratio 
of their diameters, it follows that 


(sphere of diam. 1 finger-breadth) + 64,000 poppy-seeds 

> 64,000 x 10,000 

> 640,000,000 

+6 units of second | grains 
order + 40,000,000 >} of 
units of first order | sand. 

(a fortiori) « 10 units of second 

order of numbers. 


THE SAND-RECKONER. 


231 


We now gradually increase the diameter of the supposed 


sphere, multiplying it by 100 each time. 


Thus, remembering 


that the sphere is thereby multiplied by 100° or 1,000,000, the 
number of grains of sand which would be contained in a sphere 
with each successive diameter may be arrived at as follows. 


Diameter of sphere. 


(1) 100 finger-breadths 


(2) 10,000 finger-breadths 


(3) 1 stadium 
(< 10,000 finger-breadths) 
(4) 100 stadia 


(5) 10,000 stadia 


(6) 1,000,000 stadia 


(7) 100,000,000 stadia 


(8) 10,000,000,000 stadia 


Corresponding number of grains of sand. 


< 1,000,000 x 10 units of second order 

<(7th term of series) x (10th term of 
series) 

< 16th term of series {i.e. 1015] 

<[107 or] 10,000,000 units of the second 
order. 

< 1,000,000 x (last number) 

< (7th term of series) x (16th term) 

< 22nd term of series [i.e. 1071] 

< [108 or] 100,000 units of third order. 

< 100,000 units of third order. 


< 1,000,000 x (last number) 

< (7th term of series) x (22nd term) 
< 28th term of series 

< [103 or] 1,000 units of fourth order. 
< 1,000,000 x (last number) 

< (7th term of series) x (28th term) 
< 34th term of series 

< 10 units of fifth order. 

< (7th term of series) x (34th term) 
< 40th term [1059] 
«- [ΠΟΥ or] 10,000,000 units of fifth order. 
< (7th term of series) x (40th term) 

< 46th term [105] 
«[108 or] 100,000 units of sixth order. 

< (7th term of series) x (46th term) 

< 52nd term of series [1057 
«[Π08 or] 1,000 units of seventh order. 


[101] 


[105] 


But, by the proposition above [p. 227], 
(diameter of ‘ universe’) < 10,000,000,000 stadia. 
Hence the number of grains of sand which could be contained 


in ὦ sphere of the size of our ‘universe’ is less than 1,000 units 
of the seventh order of numbers [or 1051]. 


232 ARCHIMEDES 


From this we can prove further that a sphere of the size 
attributed by Aristarchus to the sphere of the fixed stars would 
contain a number of grains of sand less than 10,000,000 units 
of the eighth order of numbers [or 10°+7 = 109], 

For, by hypothesis, 

(earth) : (‘ universe ’) = ( universe’) : (sphere of fixed stars). 
And [p. 227] 
(diameter of ‘ universe’) < 10,000 (diam. of earth) ; 
whence 
(diam. of sphere of fixed stars) < 10,000 (diam. of ‘ universe’). 
Therefore 
(sphere of fixed stars) < (10,000). (‘ universe’). 
It follows that the number of grains of sand which would be 
contained in a sphere equal to the sphere of the fixed stars 
< (10,000)° x 1,000 units of seventh order 
< (13th term of series) x (52nd term of series) 
< 64th term of series [1.6. 109] 
« [107 or] 10,000,000 units of eighth order of numbers. 


Conclusion. 


“T conceive that these things, king Gelon, will appear 
incredible to the great majority of people who have not studied 
mathematics, but that to those who are conversant therewith 
and have given thought to the question of the distances and 
sizes of the earth the sun and moon and the whole universe the 
proof will carry conviction. And it was for this reason that 
I thought the subject would be not inappropriate for your 
consideration.” 


QUADRATURE OF THE PARABOLA. 


“ ARCHIMEDES to Dositheus greeting. 


“When I heard that Conon, who was my friend in his life- 
time, was dead, but that you were acquainted with Conon and 
withal versed in geometry, while I grieved for the loss not only 
of a friend but of an admirable mathematician, I set myself the 
task of communicating to you, as I had intended to send to 
Conon, a certain geometrical theorem which had not been 
investigated before but has now been investigated by me, and 
which I first discovered by means of mechanics and then 
exhibited by means of geometry. Now some of the earlier 
geometers tried to prove it possible to find a rectilineal area 
equal to a given circle and a given segment of a circle; and 
after that they endeavoured to square the area bounded by the 
section of the whole cone* and a straight line, assuming lemmas 
not easily conceded, so that it was recognised by most people 
that the problem was not solved. But I am not aware that 
any one of my predecessors has attempted to square the 
segment bounded by a straight line and a section of a right- 
angled cone [a parabola], of which problem I have now dis- 
covered the solution. For it is here shown that every segment 
bounded by a straight line and a section of a right-angled cone 
[a parabola] is four-thirds of the triangle which has the same base 
and equal height with the segment, and for the demonstration 

* There appears to be some corruption here: the expression in the text is 
τᾶς ὅλου τοῦ κώνου τομᾶς, and it is not easy to give a natural and intelligible 
meaning to it. The section of ‘the whole cone’ might perhaps mean a section 
cutting right through it, i.e. an ellipse, and the ‘ straight line’ might be an axis 
or a diameter. But Heiberg objects to the suggestion to read τᾶς ὀξυγωνίου 
κώνου τομᾶς, in view of the addition of καὶ εὐθείας, on the ground that the former 


expression always signifies the whole of an ellipse, never a segment of it 
(Quaestiones Archimedeae, p. 149). 


234, ARCHIMEDES 


of this property the following lemma is assumed: that the 
excess by which the greater of (two) unequal areas exceeds 
the less can, by being added to itself, be made to exceed any 
given finite area. The earlier geometers have also used this 
lemma ; for it is by the use of this same lemma that they have 
shown that circles are to one another in the duplicate ratio of 
their diameters, and that spheres are to one another in the 
triplicate ratio of their diameters, and further that every 
pyramid is one third part of the prism which has the same base 
with the pyramid and equal height; also, that every cone is 
one third part of the cylinder having the same base as the cone 
and equal height they proved by assuming a certain lemma 
similar to that aforesaid. And, in the result, each of the afore- 
said theorems has been accepted* no less than those proved 
without the lemma. As therefore my work now published has 
satisfied the same test as the propositions referred to, I have 
written out the proof and send it to you, first as investigated 
by means of mechanics, and afterwards too as demonstrated by 
geometry. Prefixed are, also, the elementary propositions in 
conics which are of service in the proof (στουχεῖα κωνικὰ χρεῖαν 
ἔχοντα és τὰν ἀπόδειξιν). Farewell.” 


Proposition 1. 


If from a point on a para- 
bola a straight line be drawn 3 
which is either itself the axis or 
parallel to the axis, as PV, and 
uf QQ’ be a chord parallel to 
the tangent to the parabola at P P ν 
and meeting PV in V, then 


OV =VQ. 
Conversely, if QV = VQ’, the 
chord QQ’ will be parallel to the 
tangent at P. 


* The Greek of this passage is: συμβαίνει δὲ τῶν προειρημένων θεωρημάτων 
ἕκαστον μηδὲν ἧσσον τῶν ἄνευ τούτου Tov λήμματος ἀποδεδειγμένων πεπιστευκέναι. 
Here it would seem that πεπιστευκέναι must be wrong and that the passive 
should have been used. 


QUADRATURE OF THE PARABOLA. 235 


Proposition 2. 


Tf in a parabola QQ’ be a chord parallel to the tangent at P, 
and if a straight line be drawn through P which 1s either rtself 
the axis or parallel to the aais, and which meets QQ’ in V and 
the tangent at Q to the parabola in T, then 


a eae) Pa 


Proposition S. 


If from a point on a parabola a straight line be drawn 
which is either itself the axis or parallel to the axis, as PV, 
and if from two other points Q, Q’ on the parabola straight 
lines be drawn parallel to the tangent at P and meeting PV in 
V, V’ respectively, then 

PV PV SOV OV 2 


“ And these propositions are proved in the elements of conics.*” 


Proposition 4. 


If Qq be the base of any segment of a parabola, and P the 
vertex of the segment, and vf the diameter through any other point 
R meet Qq in O and QP (produced if necessary) in F, then 


OV V0 OF FR; 
Draw the ordinate RW to PV, meeting QP in Kk. 


* i.e. in the treatises on conics by Euclid and Aristaeus. 


236 ARCHIMEDES 


Then PVG Wi OV eR Ws 
whence, by parallels, 
POI PO*: PE. 


In other words, PQ, PF, PK are in continued proportion ; 
therefore 


POH = PP: Pe 
=PQ+PF:PF+PK 


= QF: KF. 
Hence, by parallels, 


QV: VO=OF: FR. 
[It is easily seen that this equation is equivalent to a change of 
axes of coordinates from the tangent and diameter to new axes 
consisting of the chord Qq (as axis of «, say) and the diameter 
through Q (as axis of y). 
2 

For, if QV=a, PV= = where p is the parameter of the 
ordinates to PV. 

Thus, if QO = a, and RO = y, the above result gives 


ως OF 
v—-a OF-y’ 
4: 
whence acy, meee ee 


or py =x (2a—2).] 


QUADRATURE OF THE PARABOLA. 237 


Proposition 5. 


If Qq be the base of any segment of a parabola, P the vertex 
of the segment, and PV its diameter, and if the diameter of the 
parabola through any other point R meet Qq in O and the 
tangent at ᾧ in E, then 

Q0 209 = ER: RO. 


Let the diameter through R meet QP in F. 


Then, by Prop. 4, 
OViViG= OF OER: 
Since QV = γᾷ, it follows that 
VG ξεν εν (1). 
Also, if VP meet the tangent in 1, 
PT=PYV, and therefore HF= OF. 
Accordingly, doubling the antecedents in (1), we have 
Cg 00 — OL OR: 
whence QO: 0Og=ER: RO. 


238 ARCHIMEDES 


Propositions 6, 7*. 


Suppose a lever AOB placed horizontally and supported at 
its middle point O. Let a triangle BCD in which the angle C is 
right or obtuse be suspended from B and Ὁ, so that C 1s attached 
to O and CD is in the same vertical line with O. Then, if P be 
such an area as, when suspended from A, will keep the system i 
equilibrium, 

P={ABCD. ) 

Take a point # on OB such that BE = 20", and draw EFH 
parallel to OCD meeting BC, BD in F, H respectively. Let G 
be the middle point of FH. 


A ie) E 


Then G is the centre of gravity of the triangle BCD. 


Hence, if the angular points B, C be set free and the 
triangle be suspended by attaching F to £, the triangle will 
hang in the same position as before, because EFG is a vertical 
straight line. “For this is proved +.” 


Therefore, as before, there will be equilibrium. 


Thus P:ABCD=OE: AO 
=e 
or P=1ABCD. 


* In Prop. 6 Archimedes takes the separate case in which the angle BCD of 
the triangle is a right angle so that C coincides with O in the figure and F with 
E, He then proves, in Prop. 7, the same property for the triangle in which 
BCD is an obtuse angle, by treating the triangle as the difference between two 
right-angled triangles BOD, BOC and using the result of Prop. 6. I have com- 
bined the two propositions in one proof, for the sake of brevity. The same 
remark applies to the propositions following Props. 6, 7. 

+ Doubtless in the lost book περὶ ζυγῶν. Cf. the Introduction, Chapter IL., 
ad jin. 


QUADRATURE OF THE PARABOLA. 239 


Propositions 8, 9. 


Suppose a lever AOB placed horizontally and supported at 
its middle point O. Let a triangle BCD, right-angled or obtuse- 
angled at C, be suspended from the points B, EK on OB, the 
angular point C being so attached to EK that the side CD is in the 
same vertical line with E. Let Q be an area such that 

AO: 0H=ABCD: ἢ: 
Then, if an area P suspended from A keep the system in 
equilibrium, 
P< ABCD but > Q. 

Take G the centre of gravity of the triangle BCD, and draw 

GH parallel to DC, i.e. vertically, meeting BO in H. 


A OME H B 


We may now suppose the triangle BCD suspended from H, 
and, since there is equilibrium, 


ESB CDE MANOR OUT 5, τ vate sons (i); 
whence P<eABCD. 
Also ABCD :Q=A0: OZ. 
Therefore, by (1), ABCD:Q>ABCD: P, 
and PQ. 


Propositions 10, 11. 


Suppose a lever AOB placed horizontally and supported at O, 
its middle point. Let CDEF be a trapezium which can be so 
placed that its parallel sides CD, FE are vertical, while C ds 
vertically below O, and the other sides CF, DE meet in B. Let 
EF meet BO in H, and let the trapezium be suspended by attaching 
F to H and Cto O. Further, suppose Q to be an area such that 


AO : OH = (trapezium CDEF) : Q. 


240 ARCHIMEDES 


Then, if P be the area which, when suspended from A, keeps the 


system in equilibrium, 


Pee: 
The same is true in the particular case where the angles at 
C, F are right, and consequently C, F coincide with O, H 
respectively. 
Divide OH in K so that 
(20D + FE) :(2FE+CD)=HK : KO. 


A 


Draw KG parallel to OD, and let G be the middle point of 
the portion of AG intercepted within the trapezium. Then G 
is the centre of gravity of the trapezium [On the equilibrium of 
planes, I. 15]. 

Thus we may suppose the trapezium suspended from K, and 
the equilibrium will remain undisturbed. 

Therefore 

AO: OK = (trapezium CDEF): P, 
and, by hypothesis, 
AO: OH = (trapezium CDEF) : Q. 
Since OK < OH, it follows that 
PQ): 


Propositions 12, 13. 

If the trapezium CDEF be placed as in the last propositions, 
except that CD is vertically below a point L on OB instead of 
being below O, and the trapezium is suspended from L, H, 
suppose that Q, R are areas such that 

AO : OH = (trapezium CDEF) : Q, 
and AO: OL = (trapezium CDEF) : R. 


QUADRATURE OF THE PARABOLA. 241 


If then an area P suspended from A keep the system in 
equilibrium, 
PS he bute < Q: 
Take the centre of gravity G of the trapezium, as in the 
last propositions, and let the line through @ parallel to DC 
meet OB in K. 


A oO LK H B 


Then we may suppose the trapezium suspended from K, 
and there will still be equilibrium. 


Therefore (trapezium CDEF): P= AO: OK. 
Hence 
(trapezium CDEP) : P > (trapezium CDEF) : Q, 
but < (trapezium CDEF) : R. 
It follows that P=@Q but > &. 


Propositions 14, 15. 


Let Qq be the base of any segment of a parabola. Then, if 
two lines be drawn from Q, q, each parallel to the axis of the 
parabola and on the same side of Qq as the segment is, either 
(1) the angles so formed at Q, q are both right angles, or 
(2) one is acute and the other obtuse. In the latter case let 
the angle at q be the obtuse angle. 


Divide Qq into any number of equal parts at the points 
O,, O2,... On. Draw through gq, 0,, O,,... On diameters of the 
parabola meeting the tangent at Qin 2, Z,, #,,... Ε΄, and the 
parabola itself m g, R,, R,,...R,. Join OR,, QR,; ... QRu 
meeting G2, Onis, Oia One fs el, Fy, PB 00 Prat 

HA 16 


242 ARCHIMEDES 


Let the diameters Hg, ,0,,... H,O, meet a straight line 
QOA drawn through Q perpendicular to the diameters in the 
points O, H,, H,, ... H, respectively. (In the particular case 
where Qq is itself perpendicular to the diameters q will coincide 
with O, O, with H,, and so on.) 


It is required to prove that 
(1) A£qQ<3(sum of trapezia FO,, F,02,...Pn1~0,andA £,,0,Q), 
(2) ALqQ >3(sumof trapezia R,0,, R,O;,...R,,~O,and A R,0,Q). 


Suppose AO made equal to OQ, and conceive QOA as a 
lever placed horizontally and supported at O. Suppose the 
triangle HqQ suspended from OQ in the position drawn, and 
suppose that the trapezium ΜΚ, in the position drawn is 
balanced by an area P, suspended from A, the trapezium £0, 
in the position drawn is balanced by the area P, suspended 


QUADRATURE OF THE PARABOLA. 243 
from A, and so on, the triangle #,,0,Q being in like manner 
balanced by Pri. 


Then P,+P,+...+Pni will balance the whole triangle 
EqQ as drawn, and therefore 


PP we a tA BQ: [Props 6; 7] 
Again AO OH. = O00 > OH, 
= Qq : 90; 
= E,0, : OR, [by means of Prop. 5] 


= (trapezium HO,): (trapezium FO,); 
whence [Props. 10, 11] 


(Oi) >eP%. 
Next AO; OF ΞΞ 0, ΟΣ 
Ξ τ: (Les a). waco conc ses bes (a), 
while AQ ΟΞ ἡ ΣΝ 
=i Oban HO) iret atts, cabs (B); 


and, since (a) and (8) are simultaneously true, we have, by 
Props. 12, 13, 
(Op Ps E0,). 


Similarly it may be proved that 
(F,0;) > 2. => (8,0;), 
and so on. 
Lastly [Props. 8, 9] 
AE OO ΡΞ 8020. 
By addition, we obtain 
(1) (FO,)+(410,)+...+(Pn+On)+ ΔῊ, ΟΞ ΡΊΞΕ. Ἐς Εν 
»έδδᾳρ, 

or A EqQ <3 (FO, + F,0, +... + Fr On + A E,0nQ). 
(2) (R,0,)+(R,0;)+...+(Rn+10n)+ARn0O,Q<P.+P,+...+Prais 

<P,\+ Pot ...4+ Pai, a fortiori, 

«ἐδῆᾳρῳ, 
or A EqQ > 3 (RO, + R,0;+... + Rp On + A RnOnQ). 

16—2 


24.4. ARCHIMEDES 


Proposition 16. 


Suppose Qq to be the base of a parabolic segment, q being 
not more distant than Q from the vertex of the parabola. Draw 
through q the straight line qE parallel to the axis of the parabola 
to meet the tangent at Qin EF. It is required to prove that 


(area of segment)=1 AEqQ. 


For, if not, the area of the segment must be either greater 
or less than + AXqQ. 


I. Suppose the area of the 
segment greater than 1 A EqQ. 
Then the excess can, if con- 
tinually added to itself, be 
made to exceed AXgQ. And 
it is possible to find a submul- 
tiple of the triangle HgQ less 
than the said excess of the 
segment over + AXgQ. 


Let the triangle FgQ be such 
a submultiple of the triangle 
EqQ. Divide Eq into equal 
parts each equal to qf, and let 
all the points of division in- 
cluding F' be joined to Q meet- 
ing the parabola in f,, Ro, ... 
R, respectively. Through R,, R,, ... R, draw diameters of the 
parabola meeting σῷ in O,, Ος, ... O, respectively. 


Let O,R, meet QR, in F,. 
Let O,R, meet QR, in D, and QR, in F,. 
Let O,R, meet YR, in D, and QR, in F,, and so on. 


We have, by hypothesis, 
A FqQ < (area of segment) — 1A £qQ, 
or (area of segment)— AFgQ >1 AE@Q ......... (a). 


QUADRATURE OF THE PARABOLA. 245 


Now, since all the parts of gH, as gF and the rest, are equal, 
O,R, = R,F,, 0,D, = DR, = R.F,, and so on; therefore 
A FqQ =(FO, + R,0,+ D0; +...) 


i" =(FO,+ FD, + FD, +... +FraDnit+A EnRnQ)...(8). 
ut 


(area of segment) < (FO, + F,0, +... ἘΠ, +A L,0,Q). 
Subtracting, we have 
(area of segment) —A FgQ < (2,0, + B,0;+4+... 


213 Tan +A R,,0,Q), 
whence, ὦ forteorr, by (a), 


LA EqQ < (2,0, + 2,0; +... + Rp On + A R,OnQ). 
But this is impossible, since [Props. 14, 15] 
1A EqQ > (#0, + BL0;+...+ Rp On+ AR,0,Q). 
Therefore 
(area of segment) 1A HqQ. 
II. If possible, suppose the area of the segment less than 
5A EQ. 
Take a submultiple of the triangle HqQ, as the triangle 
FqQ, less than the excess of A HqQ over the area of the 
segment, and make the same construction as before. 


Since AFqQ<1ALqQ—(area of segment), 
it follows that 
A FqQ + (area of segment) < $A HqQ 
< (FO, + FO, +... + Fa1On + AEn0nQ). 
[Props. 14, 15] 
Subtracting from each side the area of the segment, we have 
A FqQ < (sum of spaces gFR,, RF R2, ... EnRnQ) 
< (PO, + FD, +... + FriDnit ΔΑ, ΠΟ), α fortiori ; 
which is impossible, because, by (8) above, 
A FqQ = FO, + FD, +... + FPaDniat AEnRnQ. 
Hence (area of segment) ¢ LA HqQ. 
Since then the area of the segment is neither less nor 
greater than +A EqQ, it is equal-to it. 


246 ARCHIMEDES 


Proposition 17. 


It is now manifest that the area of any segment of a 
parabola is four-thirds of the triangle which has the same base 
as the segment and equal height. 


Let Qq be the base of the segment, P its vertex. Then 
PQg is the inscribed triangle with the 


same base as the segment and equal 2 


ν 
height. 
q 

Since P is the vertex* of the seg- 
ment, the diameter through P bisects 
Qq. Let V be the point of bisection. 

Let VP, and g# drawn parallel to 
it, meet the tangent at Qin 7, £ re- a 
spectively. 

Then, by parallels, 

gh =2VT, 

and JE led [Prop. 2] 
so that Vi 2 Vv. E 


Hence AXqQ =4APQq. 
But, by Prop. 16, the area of the segment is equal to tA HqQ. 
Therefore (area of segment) = 4A PQq. 


Der. “In segments bounded by a straight line and any 
curve I call the straight line the base, and the height the 
greatest perpendicular drawn from the curve to the base of the 
segment, and the vertex the point from which the greatest 
perpendicular is drawn.” 


* Tt is curious that Archimedes uses the terms base and vertex of a segment 
here, but gives the definition of them later (at the end of the proposition), 
Moreover he assumes the converse of the property proved in Prop. 18. 


QUADRATURE OF THE PARABOLA. 247 


Proposition 18. 
Tf Qq be the base of a segment of a parabola, and V the 
middle point of Qq, and tf the diameter through V meet the 
curve in P, then P ws the vertex of the segment. 


Q 


τὴ 


For Qq is parallel to the tangent at P [Prop. 1]. Therefore, 
of all the perpendiculars which can be drawn from points on the 
segment to the base ἕῳ, that from P is the greatest. Hence, 
by the definition, P is the vertex of the segment. 


Proposition 19. 

If Qq be a chord of a parabola bisected in V by the diameter 
PV, and if RM be a diameter bisecting QV in M, and RW 
be the ordinate from R to PV, then 

PV =4RM. 
Q 


q 
For, by the property of the parabola, 
Vere Ors ΤῊ: 
a es 8: 
so that PV =P W,. 
whence PV =4kRM. 


248 ARCHIMEDES 


Proposition 20. 


If Qq be the base, and P the vertex, of a parabolic segment, 
then the triangle PQq is greater than half the segment PQq. 


For the chord Qq is parallel to the tangent at P, and the 
triangle PQq is half the parallelogram Qa 
formed by Qq, the tangent at P, and the 
diameters through Ὁ, q. 


Therefore the triangle PQq is greater 
than half the segment. P ν 


Cor. It follows that τέ a possible 
to inscribe in the segment a polygon such 
that the segments left over are together 
less than any assigned area. 


Proposition 21. 


If Qq be the base, and P the vertex, of any parabolic 
segment, and if R be the vertex of the segment cut off by PQ, 
then 

ἌΡ ΞΕ ΑΕ ΡΣ 

The diameter through R will bisect the chord PQ, and 
therefore also QV, where PV is the 
diameter bisecting Qg. Let the dia- 
meter through & bisect PQ in Y and 
QVin M. Join PM. 


By Prop. 19, 
PV =<RM. 
Also PVi=2Y M. 
Therefore YM —=2RY, 
and AOPOM IAP RO. 
Hence APOV =4ANPRQ, 


and A PQq =S8APRYQ. 


QUADRATURE OF THE PARABOLA. 249 
Also, if RW, the ordinate from R to PV, be produced to 
meet the curve again in 7, 
RW=rW, 
and the same proof shows that 
APQg=S8A-Pr¢. 


Proposition 22. 


If there be a series of areas A, B, Οὐ, 1), ... each of which is 
Sour times the next in order, and if the largest, A, be equal to the 
triangle PQ inscribed in a parabolic segment PQq and having 
the same base with it and equal height, then 

(A+B+C+D+...)< (area of segment PQq). 


For, since APQq =8APRQ =8APgqr, where R, 7 are the 
vertices of the segments cut off by PQ, 
Pq, as in the last proposition, 

A PQ =4(APQR + APqr). 

Therefore, since A PQq = A, 

APQR+ APqr= B. 

In lke manner we prove that the 
triangles similarly inscribed in the re- 
maining segments are together equal to 
the area C, and so on. 


Therefore A+B+C+D+... is equal to the area of a 
certain inscribed polygon, and is therefore less than the area of 
the segment. 


Proposition 23. 


Given a series of areas A, B,C, D,... Z, of which A is the 
greatest, and each is equal to four times the next in order, then 


A+B+C+...4274+34Z7=4A. 


250 ARCHIMEDES 


Take areas 0, c, αἰ, ... such that 


= 4B, 
c=10C, 

d =1D, and so on. 
Then, since b= 4B, 
and B=}A, 
Bab = +A 
Similarly C+c=4B. 

ἜΝ meena 


B+C+D+...4+Z4+b+ce+d4+...4¢2=1(A+B4+C+.. 


But = b+ce+d+...4¢y=1(B+C0+D+4...4+ Y). 


13 
“κ᾽. Ὁ 


Therefore, by subtraction, 
B+04+D+4+...4Z+2=1A4 
or A+B+C+...444+4Z=4$A4. 


ΤῊ 


QUADRATURE OF THE PARABOLA. 251 


| The algebraical equivalent of this result is of course 
Ie EG) =e 3 
e ie ( 1 Ng | 


I=] 


Proposition 24. 


Every segment bounded by a parabola and a chord Qq is 
equal to four-thirds of the triangle which has the same base as 
the segment and equal height. 


Suppose K=$APQq, 


where P is the vertex of the segment; and we have then to 

prove that the area of the segment is 

equal to K. - 
For, if the segment be not equal to 

Κ΄, it must either be greater or less. 


I. Suppose the area of the segment 
greater than K. 

If then we inscribe in the segments 
cut off by PQ, Pq triangles which have 
the same base and equal height, Le. 
triangles with the same vertices R, r as 
those of the segments, and if in the 
remaining segments we inscribe triangles in the same manner, 
and so on, we shall finally have segments remaining whose sum 
is less than the area by which the segment PQq exceeds K. 


Therefore the polygon so formed must be greater than the 
area K ; which is impossible, since [ Prop. 23] 


A+B+C+4...42Z< 4A, 
where A=/< Pio, 


Thus the area of the segment cannot be greater than Δ. 


II. Suppose, if possible, that the area of the segment is 
less than ΚΑ. 


252 ARCHIMEDES, 


If then APQqg=A, B=14A, C=1B, and so on, until we 
arrive at an area X such that X is less than the difference 
between K and the segment, we have 


A+B+C+4+...4X+1X =4A [Prop. 23] 
we 


Now, since K exceeds 4+B+C+...+X by an area less 
than X,and the area of the segment by an area greater than X, 
it follows that 


A+B+C+...+X >(the segment) ; 
which is impossible, by Prop. 22 above. 
Hence the segment is not less than [΄. 
Thus, since the segment is neither greater nor less than K, 


(area of segment PQq) = ἡ =4APQq. 


ON FLOATING BODIES. 


BOOK 1. 


Postulate 1. 


“Let it be supposed that a fluid is of such a character that, 
its parts lying evenly and being continuous, that part which 15 
thrust the less is driven along by that which is thrust the 
more; and that each of its parts is thrust by the fluid which is 
above it in a perpendicular direction if the fluid be sunk in 
anything and compressed by anything else.” 


Proposition 1. 


If a surface be cut by a plane always passing through a 
certain point, and if the section be always a circumference [of a 
circle] whose centre is the aforesaid point, the surface is that of 
a sphere. 


For, if not, there will be some two lines drawn from the 
point to the surface which are not equal. 


Suppose O to be the fixed point, and A, B to be two points 
on the surface such that OA, OB are unequal. Let the surface 
be cut by a plane passing through OA, OB. Then the section 
is, by hypothesis, a circle whose centre is O. 


Thus O0A=OB; which is contrary to the assumption. 
Therefore the surface cannot but be a sphere. 


254 ARCHIMEDES 


Proposition 2. 


The surface of any fluid at rest is the surface of a sphere 
whose centre is the same as that of the earth. 

Suppose the surface of the fluid cut by a plane through 0, 
the centre of the earth, in the curve A BCD. 

ABCD shall be the circumference of a circle. 

For, if not, some of the lines drawn from O to the curve 
will be unequal. Take one of them, OB, such that OB is 
greater than some of the lines from O to the curve and less 
than others. Draw a circle with OB as radius. Let it be HBF, 
which will therefore fall partly within and partly without the 
surface of the fluid. 


EA 5 Oo DF 


Draw OGH making with OB an angle equal to the angle 
EOB, and meeting the surface in H and the circle in G. Draw 
also in the plane an arc of a circle PQR with centre O and 
within the fluid. 


Then the parts of the fluid along PQR are uniform and 
continuous, and the part PQ is compressed by the part between 
it and AB, while the part QR is compressed by the part 
between QR and BH. Therefore the parts along PQ, QR will 
be unequally compressed, and the part which is compressed the 
less will be set in motion by that which is compressed the 
more. 

Therefore there will not be rest; which is contrary to the 
hypothesis. 


Hence the section of the surface will be the circumference 
of a circle whose centre is O; and so will all other sections by 
planes through Ὁ. 


Therefore the surface is that of a sphere with centre O. 


ON FLOATING BODIES I. 255 


Proposition 3. 


Of solids those which, size for size, are of equal weight with 
a flud will, of let down into the fluid, be immersed so that they 
do not project above the surface but do not sink lower. 


If possible, let a certain solid HF HG of equal weight, 
volume for volume, with the fluid remain immersed in it so 
that part of it, HBCF, projects above the surface. 


Draw through 0, the centre of the earth, and through the 
solid a plane cutting the surface of the fluid in the circle 


ABCD. 


Conceive a pyramid with vertex O and base a parallelogram 
at the surface of the fluid, such that it includes the immersed 
portion of the solid. Let this pyramid be cut by the plane of 
ABCD in OL, OM. Also let a sphere within the fluid and 
below GH be described with centre O, and let the plane of 
ABCD cut this sphere in PQR. 


Conceive also another pyramid in the fluid with vertex Ὁ, 
continuous with the former pyramid and equal and similar to 


it, Let the pyramid so described be cut in OM, ON by the 
plane of ABCD. 


Lastly, let 1 ΟΥ be a part of the fluid within the second 


pyramid equal and similar to the part BG'HC of the solid, and 
let SV be at the surface of the fluid. 


Then the pressures on PQ, QR are unequal, that on PQ 
being the greater. Hence the part at QR will be set in motion 


256 ARCHIMEDES 


by that at PQ, and the fluid will not be at rest; which is 
contrary to the hypothesis. 


Therefore the solid will not stand out above the surface. 


Nor will it sink further, because all the parts of the fluid 
will be under the same pressure. 


Proposition 4. 


A solid lighter than a fluid will, if immersed in tt, not be 
completely submerged, but part of it will project above the 
surface. 


In this case, after the manner of the previous proposition, 
we assume the solid, if possible, to be completely submerged and 
the fluid to be at rest in that position, and we conceive (1) a 
pyramid with its vertex at O, the centre of the earth, including 
the solid, (2) another pyramid continuous with the former and 
equal and similar to it, with the same vertex O, (3) a portion of 
the fluid within this latter pyramid equal to the immersed solid 
in the other pyramid, (4) a sphere with centre O whose surface 
is below the immersed solid and the part of the fluid in the 
second pyramid corresponding thereto. We suppose a plane to 
be drawn through the centre O cutting the surface of the 
fluid in the circle ABC, the solid in S, the first pyramid in OA, 
OB, the second pyramid in OB, OC, the portion of the fluid in 
the second pyramid in KX, and the inner sphere in PQR. 

Then the pressures on the parts of the fluid at PQ, QR are 
unequal, since S is lighter than K. Hence there will not be 
rest ; which is contrary to the hypothesis. 


Therefore the solid S cannot, in a condition of rest, οἶσθ 
completely submerged. 


ON FLOATING BODIES 1. 257 


Proposition 5. 


Any solid lighter than a fluid will, if placed in the fluid, 
be so far immersed that the weight of the solid will be equal to 
the weight of the fluid displaced. 


For let the solid be EGHF, and let BGHC be the portion 
of it immersed when the fluid is at rest. As in Prop. 3, 
conceive a pyramid with vertex O including the solid, and 
another pyramid with the same vertex continuous with the 
former and equal and similar to it. Suppose a portion of the 
fluid ST’UV at the base of the second pyramid to be equal and 
similar to the immersed portion of the solid; and let the con- 
struction be the same as in Prop. 3. 


Then, since the pressure on the parts of the fluid at PQ, QR 
must be equal in order that the fluid may be at rest, it follows 
that the weight of the portion STUV of the fluid must be 
equal to the weight of the solid HGH. And the former is 
equal to the weight of the fluid displaced by the immersed 
portion of the solid BGHC. 


Proposition 6. 


If a solid lighter than a fluid be forcibly immersed in it, the 
solid will be driven upwards by a force equal to the difference 
between its weight and the weight of the fluid displaced. 


For let A be completely immersed in the fluid, and let G 
represent the weight of A, and (G+ H) the weight of an equal 
volume of the fluid. Take a solid D, whose weight is H 

H, A. 17 


258 ARCHIMEDES 


and add it to A. Then the weight of (A + D) is less than 
that of an equal volume of the fluid; and, if (A+) 18 
immersed in the fluid, it will project so that its weight will 
be equal to the weight of the fluid displaced. But its weight 
is (G+ Hf). 


Therefore the weight of the fluid displaced is (G+), and 
hence the volume of the fluid displaced is the volume of the 
solid A. There will accordingly be rest with A immersed 
and D projecting. 

Thus the weight of D balances the upward force exerted by 
the fluid on A, and therefore the latter force is equal to H, 
which is the difference between the weight of A and the weight 
of the fluid which A displaces. 


Proposition 7. 


A solid heavier than a fluid will, if placed in it, descend 
to the bottom of the fluid, and the solid will, when weighed 
in the fluid, be lighter than its true weight by the weight of the. 
fluid displaced. 


(1) The first part of the proposition is obvious, since the 
part of the fluid under the solid will be under greater pressure, 
and therefore the other parts will give way until the solid 
reaches the bottom. 


(2) Let A be a solid heavier than the same volume of the 
fluid, and let (G+) represent its weight, while G represents 
the weight of the same volume of the fluid. 


ON FLOATING BODIES I. 259 


Take a solid B lighter than the same volume of the fluid, 
and such that the weight of B is G, while the weight of the 
same volume of the fluid is (G+ /). 


Let A and B be now combined into one solid and immersed. 
Then, since (4 + B) will be of the same weight as the same 
volume of fluid, both weights being equal to (G+H)+G, it 
follows that (4 +B) will remain stationary in the fluid. 


Therefore the force which causes A by itself to sink must 
be equal to the upward force exerted by the fluid on B by 
itself. This latter is equal to the difference between (G+) 
and G [Prop. 6]. Hence A is depressed by a force equal to 
H, i.e. its weight in the fluid is H, or the difference between 
(G+H) and 6. 


[This proposition may, I think, safely be regarded as decisive 
of the question how Archimedes determined the proportions of 
gold and silver contained in the famous crown (cf. Introduction, 
Chapter I.). The proposition suggests in fact the following 
method. 


Let W represent the weight of the crown, w, and w, the 
weights of the gold and silver in it respectively, so that 
W=w,+ ωὖ.. 


(1) Take a weight W of pure gold and weigh it in a fluid. 
The apparent loss of weight is then equal to the weight of 
the fluid displaced. If #', denote this weight, F, is thus known 
as the result of the operation of weighing. 

It follows that the weight of fluid displaced by a weight w, 


W, F 
—., 1. 


of gold is W 


17—2 


260 ARCHIMEDES 


(2) Take a weight W of pure silver and perform the same 
operation. If #, be the loss of weight when the silver is 
weighed in the fluid, we find in like manner that the weight 
of fluid displaced by οὖς is Τ ΤΣ 

(3) Lastly, weigh the crown itself in the fluid, and let F be 
the loss of weight. Therefore the weight of fluid displaced by 
the crown is Ff. 


It follows that τὶ ὦ. Ὁ Τ᾽ Fi,=F, 
or w, FP, + w,F, = (αὐ, + w,) F, 
h ἢ, ἢ, 
whence no 


This procedure corresponds pretty closely to that described 
in the poem de ponderibus et mensuris (written probably about 
500 A.D.)* purporting to explain Archimedes’ method. Ac- 
cording to the author of this poem, we first take two equal 
weights of pure gold and pure silver respectively and weigh 
them against each other when both immersed in water; this 
gives the relation between their weights in water and therefore 
between their loss of weight in water. Next we take the 
mixture of gold and silver and an equal weight of pure silver 
and weigh them against each other in water in the same 
manner. 


The other version of the method used by Archimedes is 
that given by Vitruvius}, according to which he measured 
successively the volumes of fluid displaced by three equal 
weights, (1) the crown, (2) the same weight of gold, (3) the 
same weight of silver, respectively. Thus, if as before the 
weight of the crown is W, and it contains weights w, and w, of 
gold and silver respectively, 


(1) the crown displaces a certain quantity of fluid, V say. 
(2) the weight W of gold displaces a certain volume of 


* Torelli’s Archimedes, p. 364; Hultsch, Metrol. Script. τι. 95 sq., and 
Prolegomena § 118. 
+ De architect. 1x. 3. 


ON FLOATING BODIES I. 261 


fluid, V, say; therefore a weight w, of gold displaces a volume 


τεῦ of fluid. 


(3) the weight W of silver displaces a certain volume of 
fluid, say V,; therefore a weight w, of silver displaces a volume 


oF .V, of fluid. 
40. We 
It follows that = wil + ΠΣ Vis 
whence, since W=w,+ Wy, 
Wi ἘΝ Υ,.-- Vi 
w, V—V,’ 


and this ratio is obviously equal to that before obtained, viz. 
1 ee oy | 
F-F, 

Postulate 2. 


“Let it be granted that bodies which are forced upwards in 
a fluid are forced upwards along the perpendicular [to the 
surface] which passes through their centre of gravity.” 


Proposition 8. 


If a solid in the form of a segment of a sphere, and of a 
substance lighter than a fluid, be immersed in rt so that its base 
does not touch the surface, the solid will rest in such a position 
that its axis is perpendicular to the surface ; and, if the solid be 
forced into such a position that its base touches the fluid on one 
side and be then set free, it will not remain in that position but 
will return to the symmetrical position. 


[The proof of this proposition is wanting in the Latin 


version of Tartaglia. Commandinus supplied a proof of his 
own in his edition.] 


Proposition 9. 


If a solid in the form of a segment of a sphere, and of a 
substance lighter than a fluid, be immersed in it so that tts base 
is completely below the surface, the solid will rest in such a 
position that its axis is perpendicular to the surface. 


262 ARCHIMEDES 


[The proof of this proposition has only survived in a 
mutilated form. It deals moreover with only one case out of 
three which are distinguished at the beginning, viz. that in 
which the segment is greater than a hemisphere, while figures 
only are given for the cases where the segment is equal to, or 
less than, a hemisphere. | 


Suppose, first, that the segment is greater than a hemisphere. 
Let it be cut by a plane through its axis and the centre of the 
earth ; and, if possible, let it be at rest in the position shown 
in the figure, where AB is the intersection of the plane with 
the base of the segment, DE its axis, C the centre of the 
sphere of which the segment is a part, O the centre of the 
earth. 


The centre of gravity of the portion of the segment outside 
the fluid, as F, lies on OC produced, its axis passing through ΟἹ 


Let G be the centre of gravity of the segment. Join FG, 
and produce it to H so that 

FG: GH =(volume of immersed portion) : (rest of solid). 
Join OH. 

Then the weight of the portion of the solid outside the fluid 
acts along FO, and the pressure of the fluid on the immersed 
portion along OH, while the weight of the immersed portion 
acts along HO and is by hypothesis less than the pressure of 
the fluid acting along OH. 

Hence there will not be equilibrium, but the part of the 
segment towards A will ascend and the part towards B descend, 


until DH assumes a position perpendicular to the surface of 
the fluid. 


ON FLOATING BODIES. 


BOOK II. 


Proposition 1. 


If a solid lighter than a fluid be at rest wn it, the weight of 
the solid will be to that of the same volume of the fluid as the 
immersed portion of the solid is to the whole. 

Let (A +8) be the solid, B the portion immersed in the 
fluid. 

Let (C+D) be an equal volume of the fluid, C being equal 
in volume to A and B to D. 

Further suppose the line # to represent the weight of the 
solid (A + B), (f+ ΟἽ to represent the weight of (C+D), and 
G that of D. 


Then 
weight of (A +B) : weight of (C+ D)=# : (F+G)...(1). 


264 ARCHIMEDES 


And the weight of (A+B) is equal to the weight of a 
volume 6 of the fluid [I. 5], 1.6. to the weight of D. 


That is to say, H=G. 
Hence, by (1), 
weight of (A + 8) : weight of (C+ D)=G:F+G4 
=D:0+D 
=B:A+B. 


Proposition 2. 


If a right segment of a paraboloid of revolution whose axis is 
not greater than 3 p (where p is the principal parameter of the 
generating parabola), and whose specific gravity ts less than that 
of a fluid, be placed in the fluid with its axis inclined to the 
vertical at any angle, but so that the base of the segment does not 
touch the surface of the fluid, the segment of the paraboloid will 
not remain in that position but will return to the position in 
which its axis 1s vertical. 


Let the axis of the segment of the paraboloid be AW, and 
through AW draw a plane perpendicular to the surface of the 
fluid. Let the plane intersect the paraboloid in the parabola 
BAB’, the base of the segment of the paraboloid in BB’, and 
the plane of the surface of the fluid in the chord QQ’ of the 
parabola. 


Then, since the axis AW is placed in a position not perpen- 
dicular to QQ’, BB’ will not be parallel to QQ’. 

Draw the tangent P7' to the parabola which is parallel to 
QQ’, and let P be the point of contact*. 


[From P draw PV parallel to AN meeting QQ’ in V. 
Then PV will be a diameter of the parabola, and also the 
axis of the portion of the paraboloid immersed in the fluid. 


* The rest of the proof is wanting in the version of Tartaglia, but is given 
in brackets as supplied by Commandinus. 


ON FLOATING BODIES II. 265 


Let C be the centre of gravity of the paraboloid BAB’, and 
F that of the portion immersed in the fluid. Join FC and 
produce it to H so that H is the centre of gravity of the 
remaining portion of the paraboloid above the surface. 


B’ 


Then, since AN =3AC%, 
and AN +p, 
it follows that AC} δ, 


Therefore, if CP be joined, the angle CPT is δουύθ τ. 
Hence, if CX be drawn perpendicular to PT, K will fall between 
Pand T. And,if FL, HM be drawn parallel to CK to meet 


PT, they will each be perpendicular to the surface of the fluid. 


Now the force acting on the immersed portion of the 
segment of the paraboloid will act upwards along LF, while 
the weight of the portion outside the fluid will act downwards 
along HM. 


Therefore there will not be equilibrium, but the segment 


* As the determination of the centre of gravity of a segment of a paraboloid 
which is here assumed does not appear in any extant work of Archimedes, or 
in any known work by any other Greek mathematician, it appears probable that 
it was investigated by Archimedes himself in some treatise now lost. 

+ The truth of this statement is easily proved from the property of the sub- 


normal. For, if the normal at P meet the axis in G, AG is greater than Ε 


except in the case where the normal is the normal at the vertex 4 itself. But 
the latter case is excluded here because, by hypothesis, AN is not placed vertically. 
Hence, P being a different point from 4, AG is always greater than AC; and, 
since the angle TPG is right, the angle TPC must be acute. 


266 ARCHIMEDES 


will turn so that B will rise and B’ will fall, until AW takes 
the vertical position.] 


[For purposes of comparison the trigonometrical equivalent 
of this and other propositions will be appended. 

Suppose that the angle ΝΤΡ, at which in the above figure 
the axis AW is inclined to the surface of the fluid, is denoted 
by @. 

Then the coordinates of P referred to A and the tangent 
at A as axes are 


ὍΣ Ὁ 
4 “οὐ θ, 9 cot 6, 


where p is the principal parameter. 
Suppose that AN=h, PV=k. 
Τῇ now 2’ be the distance from 7’ of the orthogonal projection 


of F on TP, and z the corresponding distance for the point C, 
we have 


epee p d 2 
x = 65 cot? 8. cos @ + 9 cot @.sin @ + 5k cos 6, 


a =" cot? 6. cos Hebe ens 6, 
4 3 
) P (oot? Ξ Ὶ 
whence «a —x=cos 0 rm (cot? 6 + 3) -- 3 (h τ 
In order that the segment of the paraboloid may turn in 
the direction of increasing the angle P7'N, α΄ must be greater 
than w, or the expression just found must be positive. 


This will always be the case, whatever be the value of 0, if 


or ΔῈ». 


Proposition 3. 


If aright segment of a paraboloid of revolution whose axis 
is not greater than 3p (where p is the parameter), and whose 
specific gravity is less than that of a fluid, be placed in the flurd 
with its axis inclined at any angle to the vertical, but so that its 


ON FLOATING BODIES II, 267 


base is entirely submerged, the solid will not remain in that posi- 
tion but will return to the position in which the axis is vertical. 

Let the axis of the paraboloid be AN, and through ΑΝ 
draw a plane perpendicular to the surface of the fluid inter- 
secting the paraboloid in the parabola BAB’, the base of the 
segment in ΒΝ Β', and the plane of the surface of the fluid in 
the chord QQ’ of the parabola. 


T MKPL 


Then, since AN, as placed, is not perpendicular to the 
surface of the fluid, QQ’ and BB’ will not be parallel. 

Draw PT parallel to QQ’ and touching the parabola at P. 
Let PT meet NA produced in 7. Draw the diameter PV 
bisecting QQ’ in V. PV is then the axis of the portion of the 
paraboloid above the surface of the fluid. 

Let C be the centre of gravity of the whole segment of the 
paraboloid, # that of the portion above the surface. Join FC 
and produce it to H so that H is the centre of gravity of 
the immersed portion. 


Then, since 46 Ὁ 5, the angle CPT is an acute angle, as in 


the last proposition. 

Hence, if CK be drawn perpendicular to PT, K will fall 
between P and 7. Also, if HM, FL be drawn parallel to Ck, 
they will be perpendicular to the surface of the fluid. 

And the force acting on the submerged portion will act 
upwards along HWM, while the weight of the rest will act 
downwards along LF produced. 

Thus the paraboloid will turn until it takes the position 
in which AW is vertical. 


268 ARCHIMEDES 


Proposition 4. 


Given a right segment of a paraboloid of revolution whose 
axis AN is greater than 2p (where p is the parameter), and 
whose specific gravity is less than that of a fluid but bears 
to it a ratio not less than (AN—ip)’: AN’, of the segment 
of the paraboloid be placed in the fluid with its amis at any 
inclination to the vertical, but so that its base does not touch 
the surface of the fluid, it will not remain in that position but 
will return to the position in which its axis rs vertical. 


Let the axis of the segment of the paraboloid be AJ, and 
let a plane be drawn through AW perpendicular to the surface 
of the fluid and intersecting the segment in the parabola BAB’, 
the base of the segment in BB’, and the surface of the fluid in 
the chord QQ’ of the parabola. 


Then AN, as placed, will not be perpendicular to QQ’. 

Draw PT parallel to QQ’ and touching the parabola at P. 
Draw the diameter PV bisecting QQ’ in V. Thus PV will be 
the axis of the submerged portion of the solid. 

Let C be the centre of gravity of the whole solid, / that of 
the immersed portion. Join FC and produce it to H so that H 
is the centre of gravity of the remaining portion. 

Now, since AN =8AC, 

and AN > 8p, 


it follows that AC 5: ; 


od 


ON FLOATING BODIES II. 269 


2 


Measure CO along CA equal to 5, and OR along OC equal to 
140. 
Then, since AN =3AC, 
and eb), 
we have, by subtraction, 
NER =3200C. 
That is, AN —-AR=300 
τ ἢ 
or AR =(AN — 3p). 


Thus (AN ὩΣ: ΝΞ πὸ: AN’, 
and therefore the ratio of the specific gravity of the solid to 
that of the fluid is, by the enunciation, not less than the ratio 
aly ἊΝ-. 

But, by Prop. 1, the former ratio is equal to the. ratio 
of the immersed portion to the whole solid, ie. to the ratio 


PV? : AN? [On Conoids and Spheroids, Prop. 24]. 


Hence PY Aine AR* > AN?, 
or ΠΕ Ale: 

It follows that 

PF (=2PV) + 24k 
+ AO. 
If, therefore, OA be drawn from O perpendicular to OA, it will 
meet PF between P and F. 

Also, if CK be joied, the triangle KCO is equal and similar 
to the triangle formed by the normal, the subnormal and the 
ordinate at P (since CO=4p or the subnormal, and KO is 
equal to the ordinate). 

Therefore CK is parallel to the normal at P, and therefore 
perpendicular to the tangent at P and to the surface of the 
fluid. 


Hence, if parallels to CK be drawn through δ΄, H, they will 
be perpendicular to the surface of the fluid, and the force 
acting on the submerged portion of the solid will act upwards 
along the former, while the weight of the other portion will 
act downwards along the latter. 


270 ARCHIMEDES 


Therefore the solid will not remain in its position but will 
turn until AV assumes a vertical position. 


[Using the same notation as before (note following Prop. 2), 
we have 


Ω 
wf -- 2 = 005 0 ἢ (cot? 8 +2) —5 (hh, 


and the minimum value of the expression within the bracket, 
for different values of 0, is 


corresponding to the position in which «4, 17 is vertical, or 0= = 
Therefore there will be stable equilibrium in that position only, 
provided that 

k & (h— 4p), 
or, if s be the ratio of the specific gravity of the solid to that of 
the fluid (= k’/h? in this case), 


s¢(h— Sp)'/l*] 


Proposition 5. 


Given a right segment of a paraboloid of revolution such that 
its axis AN is greater than *p (where p is the parameter), and 
its specific gravity is less than that of a fluid but in a ratio to 
it not greater than the ratio {AN*—(AN —3p)}: AN’, of the 
segment be placed in the fluid with its axis inclined at any angle 
to the vertical, but so that its base is completely submerged, τέ will 
not remain in that position but will return to the position im 
which AN 1s vertical. 

Let a plane be drawn through AW, as placed, perpendicular 
to the surface of the fluid and cutting the segment of the 
paraboloid in the parabola BAB’, the base of the segment in 
BB’, and the plane of the surface of the fluid in the chord 
QQ’ of the parabola. 

Draw the tangent PZ’ parallel to QQ’, and the diameter 
PY, bisecting QQ’, will accordingly be the axis of the portion 
of the paraboloid above the surface of the fluid. 


ON FLOATING BODIES II. OFT 


Let F be the centre of gravity of the portion above the 
surface, C that of the whole solid, and produce FC to H, the 
centre of gravity of the immersed portion. 


As in the last proposition, AC > 4 and we measure CO along 


CA equal to E , and OR along OC equal to $A0. 
Then AN =3A0C, and AR=3A0; 


and we derive, as before, 
AR =(AN — 8p). 
Now, by hypothesis, 
(spec. gravity of solid) : (spec. gravity of fluid) 
} {AN*— (AN — 4p} : AN? 
>(AN*—AR?’): AN’. 


Therefore 
(portion submerged) : (whole solid) 
+ (AN? — AR’): AN’, 


and (whole solid) : (portion above surface) 
sees AL 
Thus ACN eb ae AEN ae Arle, 
whence PV Act, 
and PF+¢2AR 


«40. 


272 ARCHIMEDES 


Therefore, if a perpendicular to AC be drawn from 0, it will 
meet PF in some point K between P and F. 

And, since CO = 4p, CK will be perpendicular to P7, as in 
the last proposition. 

Now the force acting on the submerged portion of the solid 
will act upwards through H, and the weight of the other 
portion downwards through F, in directions parallel in both 
cases to CK; whence the proposition follows. 


Proposition 6. 


If a right segment of a paraboloid lighter than a fluid be 
such that its axis AM is greater than 3p, but AM:4p< 15:4, 
and if the segment be placed in the fluid with its axis so inclined 
to the vertical that its base touches the fluid, τέ will never remain 
in such a position that the base touches the surface in one point 
only. 

Suppose the segment of the paraboloid to be placed in the 
position described, and let the plane through the axis AM 
perpendicular to the surface of the fluid intersect the segment 
of the paraboloid in the parabolic segment BAB’ and the plane 
of the surface of the fluid in BQ. 


Take C on AM such that AC =2CM (or so that C is the 
centre of gravity of the segment of the paraboloid), and measure 
CK along CA such that 

aM: CK=15: 4. 


͵ 


Β 


᾿ 


Thus AM:CK > AM: 4p, by hypothesis; therefore CK < tp. 


ON FLOATING BODIES II. 273 


Measure CO along CA equal to tp. Also draw KR per- 
pendicular to AC meeting the parabola in R. 


Draw the tangent PT’ parallel to BQ, and through P draw 
the diameter PV bisecting BQ in V and meeting KR in 7. 


Then Vee: AK, 
“for this 1s proved.” * 

And CK = 4AM =2A0; 
whence AK=AC—-CK =2AC=2AM. 

Thus KM =3AM. 

Therefore KM =3AK., 

It follows that 

od een £PI, 

so that PE ae ZIV. 


Let 1’ be the centre of gravity of the immersed portion of 
the paraboloid, so that PF =2FV. Produce FC to A, the 
centre of gravity of the portion above the surface. 


Draw OL perpendicular to PV. 


* We have no hint as to the work in which the proof of this proposition was 
contained. The following proof is shorter than Robertson’s (in the Appendix 
to Torelli’s edition). 


Let BQ meet AM in U, and let PN be the ordinate from P to AM. 


We have to prove that PV . AK --. PI.KM, or in other words that 
(PV .AK-PI.KM) is positive or zero. 
Now PV.AK-PI.KM=AK.PV-(AK-AN)(AM- AK) 
=AK*-AK(AM+AN-PV)+AM.AN 


=AK*-AK.UM+4AM.AN, 
(since AN= AT). 


Now UM BMi— Nene N- 
Therefore UM?:p.AM=4AN?:p.AN, 
whence UM?2=4AM.AN, 
2 

or AM. AN=S 

2 UM? 
Therefore PV.AK-PI.KM=AK?-AK.UM+ =a 

UM\?2 


and accordingly (PV. AK — PI. KM) cannot be negative. 
HEA 18 


974, ARCHIMEDES 


Then, since CO = 1p, CL must be perpendicular to PT and 
therefore to the surface of the fluid. 

And the forces acting on the immersed portion of the 
paraboloid and the portion above the surface act respectively 
upwards and downwards along lines through F and # parallel 
to CL. 

Hence the paraboloid cannot remain in the position in which 
B just touches the surface, but must turn in the direction of 
increasing the angle ΡΤ. 

The proof is the same in the case where the point J is not 
on VP but on VP produced, as in the second figure*. 


! 


B 


[With the notation used on p. 266, if the base BB’ touch 
the surface of the fluid at B, we have 
BM=BV sin 6+ PN, 
and, by the property of the parabola, 
BV? =(p+4AN) PV 
= pk (1 + cot? @). 


Therefore V ph = Vpk +5 cot 0. 


To obtain the result of the proposition, we have to eliminate 
k between this equation and 


uf — w= cos θ {Fi (cot 9. 3) --ϑ(λ πα). 


* Tt is curious that the figures given by Torelli, Nizze and Heiberg are all 
incorrect, as they all make the point which I have called I lie on BQ instead of 
VP produced. 


ON FLOATING BODIES II. 


bo 
-1 
Or 


We have, from the first equation, 
k=h—NVphcot θ.} cot’ 6, 


or h—k="/ph cot 0-7 cot? 8. 


Therefore 


vf w= e030 1B (cot? +2) —<(Vph cot 0 — cot? ay} 


= cos 9} (1 cot? 0+ 2) ~ 3 vph cot OF. 


If then the solid can never rest in the position described, 
but must turn in the direction of increasing the angle P7'M, 
the expression within the bracket must be positive whatever 
be the value of @. 

Therefore (3) ph < 8p’, 
or h<p.] 


Proposition 7. 


Given a right segment of a paraboloid of revolution lighter 
than a fluid and such that its axis AM is greater than }p, but 
AM :}p<15:4, of the segment be placed in the fluid so that 
its base is entirely submerged, it will never rest in such a position 
that the base touches the surface of the fluid at one point only. 


Suppose the solid so placed that one point of the base 
only (6B) touches the surface of the fluid. Let the plane 
through B and the axis AM cut the solid in the parabolic 
segment BAB’ and the plane of the surface of the fluid in the 
chord BQ of the parabola. 


Let C be the centre of gravity of the segment, so that 
AC=2CM; and measure CK along CA such that 


AM SO KEM, 3, 4. 
It follows that CK <p. 


Measure CO along CA equal to $p. Draw KR perpen- 
dicular to AM meeting the parabola in R. 


18—2 


276 ARCHIMEDES 


Let PT, touching at P, be the tangent to the parabola 
which is parallel to BQ, and PV the diameter bisecting BQ, Le. 
the axis of the portion of the paraboloid above the surface. 


oO 


\ 


Cc 


\ 
λ 


Then, as in the last proposition, we prove that 


PY =, $FI, 


or> 2 


and iy 977. 


Let F be the centre of gravity of the portion of the solid 
above the surface; jom FC and produce it to H, the centre of 
gravity of the portion submerged. 


Draw OL perpendicular to PV; and, as before, since 
CO=t>p, CL is perpendicular to the tangent PY. And the 
lines through H, F parallel to CZ are perpendicular to the 
surface of the fluid; thus the proposition is established as 
before. 


The proof is the same if the point J is not on VP but on 
VP produced. 


Proposition 8. 


Given a solid in the form of a right segment of a paraboloid 
of revolution whose axis AM is greater than 3p, but such that 
AM :3p<15: 4, and whose specific gravity bears to that of a 
fluid a ratio less than (AM— 3p)’: AM”, then, tf the solid be 
placed in the fluid so that its base does not touch the fluid and 
its axis is inclined at an angle to the vertical, the solid will not 
return to the position in which its axis is vertical and will not 


ON FLOATING BODIES II. 7 | 


remain in any position except that in which its axis makes with 
the surface of the flud a certain angle to be described. 


Let am be taken equal to the axis AW, and let ὁ be a point 
on am such that ac=2cm. Measure co along ca equal to 4p, 
and or along oc equal to 4 ao. 


Let X + Y be a straight line such that 
(spec. gr. of solid) : (spec. gr. of fluid) =(X + VY)’: am?...... (a), 
and suppose ἃ = 2Y. 


poe δ = Ὁ (Ὁ Ξε: ἃ 
Now ar =3a0=3(2am—4tp) 

=am— ἢ p 

ἜΣ By 

=AM—#p. 


Therefore, by hypothesis, 
(X+Y)?: am?<ar’: am’, 
whence (X + Y)< ar, and therefore X < ao. 


Measure ob along oa equal to X, and draw bd perpendicular 
to ab and of such length that 
GEA COR OD ricsceeseeseseas veesaas (ΘΝ 
Join ad. 
Now let the solid be placed in the fluid with its axis AM 
inclined at an angle to the vertical. Through AM draw a 
plane perpendicular to the surface of the fluid, and let this 


278 ARCHIMEDES 
plane cut the paraboloid in the parabola BAB’ and the plane 
of the surface of the fluid in the chord QQ’ of the parabola. 


Draw the tangent PT parallel to QQ’, touching at P, and 
let PV be the diameter bisecting QQ’ in V (or the axis of the 
immersed portion of the solid), and PV the ordinate from P. 


Measure AO along AM equal to ao, and OC along OM 
equal to oc, and draw OL perpendicular to PV. 


I. Suppose the angle O7'P greater than the angle dab. 


Thus PN NT" > db": ba’. 
But Nee Na —p: “AN. 
=co : NT, 
and db? : ba®=4co0 : ab, by (3) 
Therefore NT < 2ab, 
or AN <ab, 
whence NO>bo (since ao= AO) 
> Α΄. 


Now (X + Y)? : am’ =(spec. gr. of solid) : (spec. gr. of fluid) 
= (portion immersed) : (rest of solid) 
1 Es es fick 
so that eyes 
But Pie NO) > XxX 
>2(X+Y), smee X =279 
>2PYV, 
or ἘΞ ΤῚΣ 
and therefore 5 UY, 


Take a point F on PV so that PF=2FV, 1.6. so that Fis 
the centre of gravity of the immersed portion of the solid. 


Also AC =ac=2%am=2AM, and therefore C is the centre 
of gravity of the whole solid. 


Join FC and produce it to H, the centre of gravity of the 
portion of the solid above the surface. 


ON FLOATING BODIES II. 279 


Now, since CO =4p, CL is perpendicular to the surface of 
the fluid; therefore so are the parallels to CZ through F and 
H. But the force on the immersed portion acts upwards 
through F and that on the rest of the solid downwards 
through H. 


Therefore the solid will not rest but turn in the direction of 
diminishing the angle MT7'P. 


II. Suppose the angle O7'P less than the angle dab. In 
this case, we shall have, instead of the above results, the 
following, 


AN > ab, 

NO < X. 
Also PV Se PL, 
and therefore PL<Z2LV. 


Make PF equal to 2F'V, so that F is the centre of gravity 
of the immersed portion. 


And, proceeding as before, we prove in this case that the 
solid will turn in the direction of increasing the angle MTP. 


III. When the angle M7'P is equal to the angle dab, 
equalities replace inequalities in the results obtained, and L is 
itself the centre of gravity of the immersed portion. Thus all 
the forces act in one straight line, the perpendicular CL; 
therefore there is equilibrium, and the solid will rest in the 
position described. 


280 ARCHIMEDES 


[With the notation before used 
xz —x2=cos@ {cot @+2) -- 56.-ὴὶ fs 


and a position of equilibrium is obtained by equating to zero the 
expression within the bracket. We have then 
Pe 
4 cot? 6 3 ( 
It is easy to verify that the angle Θ᾽ satisfying this equation 
is the identical angle determined by Archimedes. For, in the 
above proposition, 


h—k)-F. 


eA In 
2 
a ee Ρ 
whence ab = «ἢ --ς -- οὔ τες (ἢ --ἰ)---. 
Also bd? =? ab. 


It follows that 
cot? dab = αὐ" δα" = ae (h—k) -- a J 


Proposition 9. 


Given a solid in the form of a right segment of a paraboloid 
of revolution whose axis AM is greater than 2p, but such that 
AM :}p<15:4, and whose specific gravity bears to that of a 
fluad a ratio greater than |AM*—(AM—32>p)*|: AM’, then, if 
the solid be placed in the fluid with its axis inclined at an angle 
to the vertical but so that its base is entirely below the surface, 
the solid will not return to the position in which its aaxis is 
vertical and will not remain in any position except that in which 
its aais makes with the surface of the fluid an angle equal to that 
described in the last proposition. 


Take am equal to AM, and take ὁ on am such that ac = 2em. 
Measure co along ca equal to $p, and ar along ac such that 
ar = 3a0. 


ON FLOATING BODIES II, 281 


Let X + Y be such a line that 
(spec. gr. of solid) : (spec. gr. of fluid) = {am? -- (ἃ + Y)*} : am’, 
and suppose X =2Y. 


Now 


Therefore, by hypothesis, 
am? —ar? : am® < jam* —(X + Y)*} : am?, 
whence Kea ar, 
and therefore X < ao. 

Make ob (measured along oa) equal to X, and draw bd 
perpendicular to ba and of such length that 

bd? =4c0 . ab. 
Join ad. 

Now suppose the solid placed as in the figure with its axis 
AM inclined to the vertical. Let the plane through AM 
perpendicular to the surface of the fluid cut the solid in the 
parabola BAB’ and the surface of the fluid in QQ’. 

Let PT be the tangent parallel to QQ’, PV the diameter 


bisecting QQ’ (or the axis of the portion of the paraboloid above 
the surface), PN the ordinate from P. 


282 ARCHIMEDES 


I. Suppose the angle M7P greater than the angle dab. 
Let AM be cut as before in C and O so that AC=2CM, 
OC =p, and accordingly AM, am are equally divided. Draw 
OL perpendicular to PV. 


Then, we have, as in the last proposition, 


iNet > ab” : ba”, 


whence co: NT >4co : ab, 
and therefore ΑΝ « αὖ. 
It follows that NO > bo 
> X. 


Again, since the specific gravity of the solid is to that of 
the fluid as the immersed portion of the solid to the whole, 


AM—(X+Y)?: AM’=AM’-PV’: AM’, 


or Cee 7 — PV" ys AM" 
That is, A+Y=PYV. 
And ΠῚ (οὐ NO) > xX 
>, 
so that PL>2LYV. 


Take F on PV so that PF=2FV. Then Fis the centre 
of gravity of the portion of the solid above the surface. 


Also C is the centre of gravity of the whole solid. Join FC 
and produce it to H, the centre of gravity of the immersed 
portion. 


Then, since CO = 3p, CL is perpendicular to PT and to the 
surface of the fluid; and the force acting on the immersed 
portion of the solid acts upwards along the parallel to OZ 
through H, while the weight of the rest of the solid acts down- 
wards along the parallel to CL through F. 


Hence the solid will not rest but turn in the direction of 


diminishing the angle M7'P. 


II. Exactly as in the last proposition, we prove that, if the 
angle MTP be less than the angle dab, the solid will not remain 


ON FLOATING BODIES II. 283 


in its position but will turn in the direction of increasing the 
angle MTP. 


B 


Ill. If the angle MTP is equal to the angle dab, the solid 
will rest in that position, because LZ and F' will coincide, and all 
the forces will act along the one line CL. 


Proposition 10. 

Given a solid in the form of a right segment of a paraboloid 
of revolution in which the axis AM is of a length such that 
AM :4p>15:4, and supposing the solid placed in a fluid 
of greater specific gravity so that its base is entirely above the 
surface of the fluid, to investigate the positions of rest. 

(Preliminary. ) 

Suppose the segment of the paraboloid to be cut by a plane 


through its axis AM in the parabolic segment BAB, of which 
BB, is the base. 


Divide AM at C so that AC = 2CWM, and measure CK along 
CA so that 
AM :CK=15: 4 
whence, by the hypothesis, CK > 1. 


Suppose CO measured along CA equal to }p, and take a 
point R on AM such that MR -- 800. 


Thus AR=AM-—MR 
=$(AC —CO) 
= 840. 


284 ARCHIMEDES 


Join BA, draw KA, perpendicular to AM meeting BA in A, 
bisect BA in A,;,and draw A,M,, A,M, parallel to AM meeting 
BM in M,, M; respectively. 


Bi 


ot 


a a ee ne 


On A,M,, A;M, as axes describe parabolic segments similar 
to the segment BAB,. (It follows, by similar triangles, that 
BM will be the base of the segment whose axis is A;M, and 
BB, the base of that whose axis is 4.}}., where BB, = 2BM,.) 

The parabola BA,b, will then pass through (Οἱ 

[For BM,: M.M= BM, : A,K 

=KM:AK 
=CM+CkK:AC-—CK 
=G+4)4u:3-4)A4M 


= MA : AC. 


ON FLOATING BODIES II. 285 


Thus C is seen to be on the parabola BA,B, by the converse 
of Prop. 4 of the Quadrature of the Parabola.] 


Also, if a perpendicular to AM be drawn from Q, it will 
meet the parabola BA,B, in two points, as Q,, P,. Let (),Q.Q3D 
be drawn through Q, parallel to AM meeting the parabolas 
BAB,, BA,M respectively in Q,, Q; and BM in D; and let 
P,P,P, be the corresponding parallel to AM through P,. Let 
the tangents to the outer parabola at P,, ᾿ς meet MA produced 
in 17᾽,, U respectively. 

Then, since the three parabolic segments are similar and 
similarly situated, with their bases in the same straight line 
and having one common extremity, and since Q,Q.Q;D 1s a 
diameter common to all three segments, it follows that 


Q,Q2 : Q.Q; = (BB, : B,B).(BM : MB,)*. 


Now BB, : B,.B= MM, : BM (dividing by 2) 
= aE by means of (8) above. 
And BM: MB,= BM : (2BM,— BM) 
=5:(6—5), by means of (), 
=5:1. 


* This result is assumed without proof, no doubt as being an easy deduction 
from Prop. 5 of the Quadrature of the Parabola. It may be established 
as follows. 

First, since 44,4,B is a straight line, and AN=AT with the ordinary 
notation (where PT is the tangent at P and PN the ordinate), it follows, by 
similar triangles, that the tangent at B to the outer parabola is a tangent to 
each of the other two parabolas at the same point B. 

Now, by the proposition quoted, if DQ;Q.Q, produced meet the tangent BT 
in Ε, 

EQ; : Q;3D=BD : DM, 


whence Ἐρ: : ED=BD: BM. 
Similarly EQ,: ED=BD: BB,, 
and EQ, : ED=BD: BB,. 


The first two proportions are equivalent to 
EQ, : ED=BD .BB,: BM. BB,, 
and EQ, : ED=BD.BM: BM. BB,. 
By subtraction, 
Q.Q03: ED=BD.MB,: BM. BB,. 
Similarly Q:9.: ED=BD.B.B, : BB,. BB. 
It follows that 
@192 : Q20;=(B,B, : BB) .(BM : MB). 


286 ARCHIMEDES 


It follows that 
QQ» : Q.Qs =2:1, 


ἮΝ QQ. = 202Qs- 
Similarly J aed ieee ee 
Also, since Mi—sCO'= Ὁ: 

AR=AM—MR 
= AM — 8p. 
(Enunciation.) 


If the segment of the paraboloid be placed in the fluid with 
its base entirely above the surface, then 
(L.) 
(spec. gr. of solid) : (spec. gr. of fluid) ¢ AR* : AM” 
[< (AM — 3p): AM"), 
the solid will rest in the position in which its axis AM is vertical; 
(IL) if 
(spec. gr. of solid) : (spec. gr. of fluid) < AR® : AM* 
but > QQ; : AM’, 
the solid will not rest with its base touching the surface of the 
fluid in one point only, but in such a position that its base does 
not touch the surface at any point and its axis makes with the 
surface an angle greater than U; 
(III. a) of 
(spec. gr. of solid) : (spec. gr. of fluid) = QQ," : AM’, 
the solid will rest and remain in the position in which the base 
touches the surface of the fluid at one point only and the axis 
makes with the surface an angle equal to U; 
(III. b) of 
(spec. gr. of solid) : (spec. gr. of fd) = P,P: AM’, 
the solid will rest with its base touching the surface of the fluid 
at one point only and with its axis inclined to the surface at an— 
angle equal to T;; 
(LY) a7, 
(spec. gr. of solid) : (spec. gr. of fluid) > P,P; : AM* 
bub ΞΘ ee 


ON FLOATING BODIES II. 287 


the solid will rest and remain in a position with its base more 
submerged ; 


(V.) 7 
(spec. gr. of solid) : (spec. gr. of fluid) < P,P’: AM’, 


the solid will rest in a position in which its axis is inclined to the 
surface of the flud at an angle less than T;, but so that the base 
does not even touch the surface at one point. 


(Proof. ) 

(I.) Since AM > 3p, and 

spec. gr. of solid) : (spec. gr. of fluid) { (AM— 2p): AM’, 
pec. § pec. δ 4" 


it follows, by Prop. 4, that the solid will be in stable equilibrium 
with its axis vertical, 


(II.) In this case 
(spec. gr. of solid) : (spec. gr. of fluid) < AR’: AM? 
but > Q,Q,;’ : AM’, 


B M Bo Βι 


Suppose the ratio of the specific gravities to be equal to 
ies AME. 
so that 1< AR but > Q,Q. 
Place P’V’ between the two parabolas BAB,, BP,Q,M equal 


288 ARCHIMEDES 
to J and parallel to AM*; and let P’V’ meet the intermediate 
parabola in F”. 
Then, by the same proof as before, we obtain 
P'F’ =2F'V’. 
Let P’T’, the tangent at P’ to the outer parabola, meet 
MA in T’, and let P’N’ be the ordinate at P’. 


Join BV’ and produce it to meet the outer parabola in Q”. 
Let OQ.P, meet P’V’ in J. 


Now, since, in two similar and similarly situated parabolic 


* Archimedes does not give the solution of this problem, but it can be 
supplied as follows. 

Let BR,Q,, BRQ, be two similar and similarly situated parabolic segments 
with their bases in the same straight line, and let BH be the common tangent 
at B. 


Suppose the problem solved, and let HRR,O, parallel,to the axes, meet the 
parabolas in R, R, and BQ, in O, making the intercept RR, equal to lL. 


Then, we have, as usual, 
ER, : HO=BO: BQ, 
— BO. BQ, BO BOs. 
and ER: EO=B0: BQ, 


=BO.BQ, : ΒΩ, ΒΩ 
By subtraction, 
RR, : EO=BO. Q,Q, : BQ,- BQ»; 


or BO.OE=1. BQ: ae which is known. 

: 1W2 
And the ratio BO: OF is known. Therefore BO?, or OH?, can be found, and 
therefore O, 


ON FLOATING BODIES II. 289 


segments with bases BM, BB, in the same straight line, BV’, BQ’ 
are drawn making the same angle with the bases, 
BV BO = BM: BB* 
= b= 2, 
so that BV ¥'Q. 
Suppose the segment of the paraboloid placed in the fluid, 
as described, with its axis inclined at an angle to the vertical, 


and with its base touching the surface at one point B only. 
Let the solid be cut by a plane through the axis and per- 


B’ 


pendicular to the surface of the fluid, and let the plane intersect 
the solid in the parabolic segment BAB’ and the plane of the 
surface of the fluid in BQ. 


Take the points ὦ, O on AM as before described. Draw 


* To prove this, suppose that, in the figure on the opposite page, BR, is 
produced to meet the outer parabola in R,. 


We have, as before, 
ER, : EO=B0: BQ, 


ER: EO=B0: BQ,, 
whence ER, : ER=BQ, : BQ). 
And, since R, is a point within the outer parabola, 
ER: ER,=BR, : BR,, in like manner. 
Hence BQ, : BQ.=BR, : BR. 
H. A. 19 


290 ARCHIMEDES 


the tangent parallel to BQ touching the parabola in P and 
meeting AM in 7; and let PV be the diameter bisecting BQ 
(i.e. the axis of the immersed portion of the solid). 
Then 
? : AM? = (spec. gr. of solid) : (spec. gr. of fluid) 
= (portion immersed) : (whole solid) 
=e SELL”, 
whence PY’ =t=PV. 
Thus the segments in the two figures, namely BP’Q’, 
BPQ, are equal and similar. 
Therefore ΠΝ = 2 PrN’. 
Also Aa A AN AN’, PN = PON 
Now, in the first figure, P’I « 21V". 
Therefore, if OZ be perpendicular to PV in the second 
figure, 
Sel <r) ON 
Take F on LV so that PF' = 2FV,1.e. so that Fis the centre 
of gravity of the immersed portion of the solid. And C is the 


centre of gravity of the whole solid. Join FC and produce it to 
H, the centre of gravity of the portion above the surface. 


Now, since CO = 4p, CL is perpendicular to the tangent at 
P and to the surface of the fluid. Thus, as before, we prove 
that the solid will not rest with B touching the surface, but will 
turn in the direction of increasing the angle P7'N. 


Hence, in the position of rest, the axis AM must make with 
the surface of the fluid an angle greater than the angle U which 
the tangent at @, makes with AM. 


(III. a) In this case 
(spec. gr. of solid) : (spec. gr. of fluid) = Q,Q,” : AM”. 


Let the segment of the paraboloid be placed in the fluid so 
that its base nowhere touches the surface of the fluid, and its 
axis 15 inclined at an angle to the vertical. 


ON FLOATING BODIES II. 291 


Let the plane through AM perpendicular to the surface of 
the fluid cut the paraboloid in the parabola BAB’ and the 


Bi 


plane of the surface of the fluid in QQ’. Let PT be the tangent 
parallel to QQ’, PV the diameter bisecting QQ’, PN the ordinate 
at P. 
Divide AWM as before at C, Ο. 
19—2 


292 ARCHIMEDES 


In the other figure let Q,N’ be the ordinate at ὧι. Join 
BQ; and produce it to meet the outer parabola in g. Then 
BQ; = Q:g, and the tangent Q,U is parallel to Bg. Now 

Q,Q.? : AM? = (spec. gr. of solid) : (spec. gr. of fluid) 
= (portion immersed) : (whole solid) 
te A. 

Therefore Q,Q; = PV; and the segments QPQ’, BQ,g of the 
paraboloid are equal in volume. And the base of one passes 
through B, while the base of the other passes through Q, a point 
nearer to A than B is. 


It follows that the angle between QQ’ and BB’ is less than 
the angle B, Bq. 


Therefore mL PEN, 
whence AN’>AN, 
and therefore πίον ΟΞ PL 


where OL is perpendicular to PV. 


It follows, since Q,Q. = 2Q.Q;, that 
PS EVA 


Therefore F, the centre of gravity of the immersed portion 
of the solid, is between P and ZL, while, as before, CZ is perpen- 
dicular to the surface of the fluid. 


Producing FC to H, the centre of gravity of the portion of 
the solid above the surface, we see that the solid must turn in 
the direction of diminishing the angle P7'N until one point B 
of the base just touches the surface of the fluid. 


When this is the case, we shall have a segment BPQ equal 
and similar to the segment BQ,g, the angle PTW will be equal 
to the angle U, and AWN will be equal to AN’. 


Hence in this case PL =2LV, and F, LZ coincide, so that F, 
C, H are all in one vertical straight line. 


Thus the paraboloid will remain in the position in which 
one point B of the base touches the surface of the fluid, and the 
axis makes with the surface an angle equal to U. 


ON FLOATING BODIES II. 293 


(III. b) In the case where 
(spec. gr. of solid) : (spec. gr. of fluid) = P,P,’ : AM’, 


we can prove in the same way that, if the solid be placed in the 
fluid so that its axis is inclined to the vertical and its base does 
not anywhere touch the surface of the fluid, the solid will take 
up and rest in the position in which one point only of the base 
touches the surface, and the axis is inclined to it at an angle 
equal to 7; (in the figure on p. 284). 


(IV.) In this case 
(spec. gr. of solid) : (spec. gr. of fluid) > P,P,’ : AM” 
but < QQ,’ : AM”. 


Suppose the ratio to be equal to 7’: AM”, so that l is greater 
than P,P, but less than Q,Q;. 


Place P’V’ between the parabolas BP,Q,, BP;Q; so that 
P’V’ is equal to ὦ and parallel to AM, and let P’V’ meet the 
intermediate parabola in F’ and OQ,P, in 1. 


Join BV’ and produce it to meet the outer parabola in gq. 


Then, as before, BV’ = V’q, and accordingly the tangent 
P’T’ at P’ is parallel to Bg. Let P’N’ be the ordinate of P’. 


294 ARCHIMEDES 


1. Now let the segment be placed in the fluid, first, with 
its axis so inclined to the vertical that its base does not 
anywhere touch the surface of the fluid. 


Let the plane through AWM perpendicular to the surface of 
the fluid cut the paraboloid in the parabola BAB’ and the 
plane of the surface of the fluid in QQ’. Let PT be the 
tangent parallel to QQ’, PV the diameter bisecting QQ’. 
Divide AWM at C, O as before, and draw OL perpendicular to PV. 


Then, as before, we have PV =1= P’V’. 


Thus the segments BP’g, QPQ’ of the paraboloid are equal 
in volume; and it follows that the angle between QQ’ and BB’ 
is less than the angle B,Bq. 


Therefore Nl ke 7 TIN: 
and hence AN’ > AWN. 
so that NO >N’O, 
Le. Piss Pal 


> P’F’, a fortiori. 
Thus PL >2LV, so that F, the centre of gravity of the 


immersed portion of the solid, is between LZ and P, while CL 
is perpendicular to the surface of the fluid. 


ON FLOATING BODIES II. 295 


If then we produce FC to H, the centre of gravity of the 
portion of the solid above the surface, we prove that the solid 
will not rest but turn in the direction of diminishing the 


angle PTN. 


2, Next let the paraboloid be so placed in the fluid that 
its base touches the surface of the fluid at one point B only, 
and let the construction proceed as before. 


Then PV =P’V’, and the segments BPQ, BP’ are equal 
and similar, so that 
LPLN= 2P Τὰν 
It follows that ΕΝ 6 I, OSI Os 
and therefore Pi PL, 
whence Pi> ΠΥ. 


Thus δ᾽ again lies between P and JZ, and, as before, the 
paraboloid will turn in the direction of diminishing the angle 
PTN, i.e. so that the base will be more submerged. 

(V.) In this case 

(spec. gr. of solid) : (spec. gr. of fluid) < P,P,’ : AM’. 
If then the ratio is equal to 7?: AM’, 1<P,P;. Place P’V’ 
between the parabolas BP,Q, and BP;Q; equal in length to / 


296 ARCHIMEDES 


and parallel to AM. Let P’V’ meet the intermediate parabola 
in δ΄ and OP, in J. 

Join BV’ and produce it to meet the outer parabola in gq. 
Then, as before, BV’ = V’g, and the tangent P’7” is parallel 
to Bq. 


B M Ba B 


‘ 


1. Let the paraboloid be so placed in the fluid that its 
base touches the surface at one point only. 


Β΄ 


ON FLOATING BODIES II, 297 


Let the plane through AM perpendicular to the surface 
of the fluid cut the paraboloid in the parabolic section BAB’ 
and the plane of the surface of the fluid in BQ. 


Making the usual construction, we find 


Ev ΞΕ VV, 
and the segments BPQ, ΒΡ. are equal and similar. 
Therefore ZEIN=Z PN, 
and aN τὴ O = NO. 
Therefore PL =P'T, 


whence it follows that PL « 31,7. 


Thus Ff, the centre of gravity of the immersed portion of the 
solid, lies between Z and V, while CZ is perpendicular to the 
surface of the fluid. 


Producing FC to H, the centre of gravity of the portion 
above the surface, we prove, as usual, that there will not be 
rest, but the solid will turn in the direction of increasing the 
angle PT'N, so that the base will not anywhere touch the 
surface. 


2. The solid will however rest in a position where its axis 
makes with the surface of the fluid an angle less than 7}. 


B’ 


298 ARCHIMEDES 


For let it be placed so that the angle PTW is not less 
than 7). 


Then, with the same construction as before, PV =1 = P’V’. 


And, since ΕΣ ΤΟ 
AN+AN,, 

and therefore NO ¢ N,O, where P,N, is the ordinate of P;. 
Hence Pe oP Ps, 
But ΡΣ» PE’. 
Therefore ΤΣ ἘΣ ΡΥ͂, 


so that F, the centre of gravity of the immersed portion of 
the solid, lies between P and L. 


Thus the solid will turn in the direction of diminishing 
the angle PTW until that angle becomes less than 7). 


[As before, if w, x’ be the distances from 7’ of the orthogonal 
projections of C, F respectively on 7'P, we have 
v! 2 = 0058 2 (cot? 0 +2)—F (hh ΕἾ τ΄. (1), 
where h= AM, k= PV. 


Also, if the base BB’ touch the surface of the fluid at one 
point B, we have further, as in the note following Prop. 6, 


Vph = Vpk +5 cot. chee (2), 


and h—k=Nphcot 0 _f cot’ 6 τ (3). 


Therefore, to find the relation between ἢ and the angle @ at 
which the axis of the paraboloid is inclined to the surface of the 
fluid in a position of equilibrium with B just touching the 
surface, we eliminate k and equate the expression in (1) to 
zero; thus 


Ὁ (cot? +2) —5 (νγῇ cot 9 -- ἢ cot! @) 0, 


4 
or 5p cot? @ — 8Vph cot 6+ 6p =0 .....ἁ(νννννν (4). 


ON FLOATING BODIES II. 299 


The two values of @ are given by the equations 
ὅν cot @=4Vh + V16h — 80} .... cece. (5). 


The lower sign corresponds to the angle U, and the upper 
sign to the angle 7, in the proposition of Archimedes, as can 
be verified thus. 


In the first figure of Archimedes (p. 284 above) we have 


AK =2h, 
M.D? =2p.0K =3p (3h —2h—4p) 
ἘΠῚ 4h -ἢ). 
hee 2 
If P, P,P; meet BM in D’, it follows that 
M,D 
uit — Med + MM, 
3p (4h -- 
=4/ 2 τῇ At τ. ἘΝ 
MD α 
and il M,D 


Now, from the property of the parabola, 


cot U=2MD/p, 
cot 7, = 2MD'/p, 
so that Feo ot tn t= 2 ph 5 4/52 (4 ἘΠ τ 
2 fa U ΠΤ Ξῃς 
or . 
1 


which agrees with the result (5) above. 


To find the corresponding ratio of the specific gravities, or 
k*/h?, we have to use equations (2) and (5) and to express k in 
terms of h and p. 


300 ARCHIMEDES 


Equation (2) gives, on the substitution in it of the value of 
cot Θ᾽ contained in (5), 


Vk =Vh—2,(4Vh + V16h — 30p) 
=3Vvh 1 t,V16h —30p, 
whence we obtain, by squaring, 
k=}38h—3,p Ὁ & νὰ (16h — 30p)......... (6). 


The lower sign corresponds to the angle U and the upper to 
the angle 7;, and, in order to verify the results of Archimedes, 
we have simply to show that the two values of ὦ are equal to 
Q,9;, PP; respectively. 


Now it is easily seen that 
Q,Q; =h/2 — MD?/p + 2M;D?/p, 
P,P; =h/2 —MD"/p + 2M;,D"/p. 
Therefore, using the values of MD, MD’, M,D, M,D' above 


found, we have 


al EG : 7h 6 veo 


P,P; 15 2) 50x 152 


which are the ne of k given in Ae above. ] 


BOOK OF LEMMAS. 


Proposition 1. 


If two circles touch at A, and if BD, EF be parallel diameters 
in them, ADF is a straight line. 

[The proof in the text only applies to the particular case 
where the diameters are perpendicular to the radius to the 
point of contact, but it is easily adapted to the more general 
case by one small change only.] 

Let O, C be the centres of the circles, and let OC be joined 
and produced to A. Draw DH parallel to AO meeting OF 


im 77. 
A 
δ. ; 
Ε 
Then, since OH =O) = CA. 
and OF —OAr 
we have, by subtraction, 


HF =CO0O = DH. 
Therefore ZHDF=2z HFD. 


302 ARCHIMEDES 


Thus both the triangles CAD, HDF are isosceles, and the 
third angles ACD, DHF in each are equal. Therefore the 


equal angles in each are equal to one another, and 
ZADC=2DFH. 
Add to each the angle CDF, and it follows that 
LADC+2CDF=2CDF +2 DFH 
= (two right angles). 
Hence ADF is a straight line. 


The same proof applies if the circles touch externally*. 


Proposition 2. 


Let AB be the diameter of a semicircle, and let the tangents 
to it. at B and at any other point D on tt meet in T. If now DE 
be drawn perpendicular to AB, and if AT, DE meet in F, 

DE = FE. 

Produce AD to meet BY produced in H. Then the angle 
ADB in the semicircle is right; therefore the angle BDH is 
also right. And 17, TD are equal. 


A E B 


Therefore 7’ is the centre of the semicircle on BH as 
diameter, which passes through D. 


Hence ἘΠ ΞΞ TB. 
And, since DE, HB are parallel, it follows that DF = FE. 


* Pappus assumes the result of this proposition in connexion with the 


ἄρβηλος (p. 214, ed. Hultsch), and he proves it for the case where the circles 
touch externally (p. 840). 


BOOK OF LEMMAS. 303 


Proposition S. 


Let P be any point on a segment of a circle whose base is 
AB, and let PN be perpendicular to AB. Take D on AB so 
that AN= ND, If now PQ be an arc equal to the arc PA, and 
BQ be joined, 


BQ, BD shall be equal*. 


8 


Α Ν D B 
Join PA, PQ, PD, DQ. 


* The segment in the figure of the ms. appears to have been a semicircle, 
though the proposition is equally true of any segment. But the case where the 
segment is a semicircle brings the proposition into close connexion with a 
proposition in Ptolemy’s μεγάλη σύνταξις, I. 9 (p. 31, ed. Halma; cf. the repro- 
duction in Cantor’s Gesch. d. Mathematik, 1. (1894), p. 389). Ptolemy’s object is 
to connect by an equation the lengths of the chord of an are and the chord of half 
the arc. Substantially his procedure is as follows. Suppose AP, PQ to be 
equal arcs, AB the diameter through 4; and let AP, PQ, AQ, PB, QB be joined. 
Measure BD along BA equal to BQ. The perpendicular PN is now drawn, and 
it is proved that PA=PD, and AN=ND. 

Then AN=}(BA-BD)=}(BA-—BQ)=}(BA-NBA?— AQ?): 

And, by similar triangles, AN TAP=AP AB. 
Therefore PAVE AB eeu 
=}(A4B-V/AB?— AQ?). AB. 
This gives AP in terms of AQ and the known diameter AB, If we divide by 


AB? throughout, it is seen at once that the proposition gives a geometrical 
proof of the formula 


The case where the segment is a semicircle recalls also the method used by 
Archimedes at the beginning of the second part of Prop. 3 of the Measurement 
of a circle. It is there proved that, in the figure above, 


AB+BQ:AQ=BP: PA, 
or, if we divide the first two terms of the proposition by AB, 


(1+ cos a)/sin a=cot δ. 


304 ARCHIMEDES 


Then, since the arcs PA, PQ are equal, 


ent), 
But, since AN = ND, and the angles at JV are right, 
ee) 
Therefore ray — sD), 
and ΖΒ ΞΕ 7 PDQ. 


Now, since A, P, Q, B are concyclic, 
ZPAD+ 2 PQB = (two right angles), 
whence ZPDA+2Z PQB = (two right angles) 


=£ PDA -Z PDE 
Therefore ZO =7 PDE: 
and, since the parts, the angles PQD, PDQ, are equal, 
ZH) = 7 BD; 
and Dy = ΒΨ: 


Proposition 4. 


If AB be the diameter of a semicircle and N any point on AB, 
and if semicircles be described within the first semicircle and 
having AN, BN as diameters respectively, the figure included 
between the circumferences of the three semicircles is “what 
Archimedes called an dpBnros*” ; and its area is equal to the 
circle on PN as diameter, where PN is perpendicular to AB 
and meets the original semicircle in P. 


For AB’? = AN’+NB*?+2AN.NB 
= AN’+ NB’ +2PN’. 


But circles (or semicircles) are to one another as the squares of 
their radu (or diameters). 


* ἄρβηλος is literally ‘a shoemaker’s knife.’ Cf. note attached to the remarks 
on the Liber Assumptorum in the Introduction, Chapter II. 


BOOK OF LEMMAS. 90 


Hence 
(semicircle on AB)=(sum of semicircles on AN, VB) 


+2 (semicircle on PJ). 


P 


A N B 


That is, the circle on PN as diameter is equal to the 
difference between the semicircle on AB and the sum of the 
semicircles on AN, ΝΒ, i.e. is equal to the area of the ἄρβηλος. 


Proposition 5. 


Let AB be the diameter of a semicircle, C any point on AB, 
and OCD perpendicular to it, and let semicircles be described 
within the first semicircle and having AC, CB as diameters. 
Then, if two circles be drawn touching CD on different sides 
and each touching two of the semicircles, the circles so drawn 
will be equal. 

Let one of the circles touch CD at 1, the semicircle on AB 
in #’, and the semicircle on AC in G. 


Draw the diameter HH of the circle, which will accordingly 
be perpendicular to CD and therefore parallel to AB. 


Join FH, HA, and FL, EB. Then, by Prop. 1, FHA, FEB 
are both straight lines, since HH, AB are parallel. 
For the same reason AGE, CGH are straight lines. 
Let AF produced meet CD in D, and let A produced 
meet the outer semicircle in J. Join BI, JD. 
H. A. 20 


306 ARCHIMEDES 


Then, since the angles AFB, ACD are right, the straight 
lines AD, AB are such that the perpendiculars on each from the 
extremity of the other meet in the point #. Therefore, by the 
properties of triangles, AH is perpendicular to the line joming 
B to D. 


A c B 


But AZ is perpendicular to BI. 
Therefore BID is a straight line. 


Now, since the angles at G, J are right, CH is parallel 
to BD. 


Therefore ΒΟ — ΠΗ͂Ι 
= AC: HE, 
so that ACCRA... HE: 
In like manner, if d is the diameter of the other circle, we can 
prove that AC. CB=AB. ἃ. 


Therefore d = HE, and the circles are equal™*. 


* The property upon which this result depends, viz. that 
AB: BC=AC: HE, 
appears as an intermediate step in a proposition of Pappus (ρ. 230, ed. Hultsch) 
which proves that, in the figure above, 
AB : BC=CE? : HE?. 
The truth of the latter proposition is easily seen. For, since the angle CEH 
is a right angle, and HG is perpendicular to CH, 
CH HHA OGG 
SANG! 9, 18105}. 


BOOK OF LEMMAS, 307 


[As pointed out by an Arabian Scholiast Alkauhi, this 
proposition may be stated more generally. If, instead of one 
point C on AB, we have two points C, D, and semicircles be 
described on AC, BD as diameters, and if, instead of the 
perpendicular to AB through C, we take the radical axis of the 
two semicircles, then the circles described on different sides of 
the radical axis and each touching it as well as two of the 
semicircles are equal. The proof is similar and presents no 


difficulty. ] 


Proposition 6. 


Let AB, the diameter of a semicircle, be divided at C so that 
AC=30CB [or in any ratio]. Describe semicircles within the 
first semicircle and on AC, CB as diameters, and suppose a 
circle drawn touching all three semicircles. If GH be the 
diameter of this circle, to find the relation between GH and AB. 


Let GH be that diameter of the circle which is parallel to 
AB, and let the circle touch the semicircles on AB, AC, CB 
in D, Μ΄, F respectively. 


Join AG, GD and BH, HD. Then, by Prop. 1, AGD, BHD 


are straight lines. 


A N Cc P B 


For a like reason AHH, BFG are straight lines, as also 
are CEG, CFH. 


Let AD meet the semicircle on AC in J, and let BD meet 
the semicircle on CB in K. Join CJ, CK meeting AL, BF 


20—2 


308 ARCHIMEDES 


respectively in Z, M, and let GZ, HM produced meet AB in 
NV, P respectively. 

Now, in the triangle AGC, the perpendiculars from dA, C on 
the opposite sides meet in LZ. Therefore, by the properties of 
triangles, GZN is perpendicular to AC. 

Similarly HWP is perpendicular to CB. 


Again, since the angles at J, K, D are right, CK is parallel 
to AD, and CI to BD. 


Therefore nCeCB=AL: LH 
=AN: NP, 

and BC :CA=BM: MG 
= BP: PN. 

Hence AN: NP=NP: PB, 


or AN, VP, PB are in continued proportion*. 
Now, in the case where AC = 3 CB, 
AN=3 NP =2 PB, 
whence BE SPM NA SAB = 49629 2 19! 
Therefore GH = NP =. AB. 
And similarly GH can be found when AC : CB is equal to 
any other given ratiot. 


* This same property appears incidentally in Pappus (p. 226) as an inter- 
mediate step in the proof of the “‘ ancient proposition ” alluded to below. 
+ In general, if AC : CB=: 1, we have 


BP: PN: NA: AB=1:2X: 22: (14+A+), 
and GH : AB=)X: (14+A+)?). 
It may be interesting to add the enunciation of the ‘‘ ancient proposition ” 
stated by Pappus (p. 208) and proved by him after several auxiliary lemmas. 


τος 
᾿ 
Α Ng No CN, 


ies) 


BOOK OF LEMMAS. 309 


Proposition 7. 


If circles be circumscribed about and inscribed in a square, 
the circumscribed circle is double of the mscribed circle. 


For the ratio of the circumscribed to the inscribed circle is 
equal to that of the square on the diagonal to the square itself, 
1.e. to the ratio 2: 1. 


Proposition 8. 


If AB be any chord of a circle whose centre is O, and if AB 
be produced to C so that BC is equal to the radius; if further CO 
meet the circle in D and be produced to meet the circle a second 
time in E, the arc AE will be equal to three tumes the are BD. 


| tee i Ge Cc 


<a 
Draw the chord EF parallel to AB, and join OB, OF. 


Let an ἄρβηλος be formed by three semicircles on AB, AC, CB as diameters, and 
let a series of circles be described, the first of which touches all three semicircles, 
while the second touches the first and two of the semicircles forming one end 
of the ἄρβηλος, the third touches the second and the same two semicircles, and 
soon, Let the diameters of the successive circles be d,, ἄς, d,... their centres 


O,, O02, O3,... and O,N,, O.N., O3Ns,... the perpendiculars from the centres on 
AB. Then it is to be proved that 


eee eee eeeeee 


310 ARCHIMEDES 


Then, since the angles OZ F, OFE are equal, 
Z2COF =2ZZ0EF 
= 2 2 BCO, by parallels, 


= 22 BOD, since BC = BO. 
Therefore 


“BOF —=3 2 BOD, 
so that the arc BF is equal to three times the are BD. 


Hence the arc AZ, which is equal to the arc BF, is equal to 
three times the are BD*. 


Proposition 9. 


If in a circle two chords AB, CD which do not pass through 
the centre intersect at right angles, then 


(arc AD)+(are CB)=(are AC) + (are DB). 


Let the chords intersect at O, and draw the diameter HF 
parallel to AB intersecting CD in 
H. EF will thus bisect CD at Ὁ 
right angles in H, and 


A i B 
(arc ED) =(are EC). 
Also EDF, ECF are semi- Η 
circles, while Ε 
(arc ED) -Ξ (ατο EA) + (are AD). 
Therefore 
(sum of arcs CF, HA, AD) = (are Ὁ 
of a semicircle). 
And the arcs AH, BF are equal. 
Therefore 
(are CB) + (are AD) = (arc of a semicircle). 

* This proposition gives a method of reducing the trisection of any angle, 
i.e. of any circular are, to a problem of the kind known as νεύσει. Suppose that 
AE isthe arc to be trisected, and that ED is the diameter through E of the circle 
of which AE is an are. In order then to find an arc equal to one-third of AE, 
we have only to draw through A a line ABC, meeting the circle again in B and 


ED produced in C, such that BC is equal to the radius of the circle. For a 
discussion of this and other νεύσεις see the Introduction, Chapter V. 


BOOK OF LEMMAS 9311] 


Hence the remainder of the circumference, the sum of the 
ares AC, DB, is also equal to a semicircle; and the proposition 
is proved. 


Proposition 10. 


Suppose that TA, TB are two tangents to a circle, while TC 
cuts it. Let BD be the chord through B parallel to TC, and let 
AD meet TC in E. Then, if EH be drawn perpendicular to BD, 
τέ will bisect it in H. 

Let AB meet TC in F, and join BE. 


Now the angle 7'AB is equal to the angle in the alternate 
segment, Le. 
ZTAB=2 ADB 


= 2 AKT, by parallels. 


Hence the triangles HAT’, AFT have one angle equal and 
another (at 7’) common. They are therefore similar, and 


“72. AA eT 


Therefore 
ἘΠ: TE =TA* 
= /M5H 
It follows that the triangles HBT, BFT are similar. 
Therefore ZTE = A TBE 
ΞΞΙ 11} 


But the angle 7.1 Β is equal to the angle 1.81), and the 
angle 7.4.8 was proved equal to the angle HDB. 


ole ARCHIMEDES 


Therefore ZEDB=Z EBD. 
And the angles at # are right angles. 
It follows that BH = HD*. 


Proposition 11. 


If two chords AB, CD in a circle intersect at right angles in 
a point O, not being the centre, then 
40" + BO’ + CO’ + DO® = (diameter)’. 
Draw the diameter CH, and join AC, CB, AD, BE. 


Then the angle CAO is equal 
to the angle CEB in the same seg- 
ment, and the angles AOC, EBC 5 ἈΞ 
are right; therefore the triangles 
AOC, EBC are similar, and 


ZACO =2 ECB. 


It follows that the subtended 
ares, and therefore the chords AD, 
BE, are equal. DE 


* The figure of this proposition curiously recalls the figure of a problem 
given by Pappus (pp. 836-8) among his lemmas to the first Book of the treatise 
of Apollonius On Contacts (περὶ ἐπαφῶν). The problem is, Given a circle and 
two points Εἰ, F (neither of which is necessarily, as in this case, the middle 
point of the chord of the circle drawn through EF, ΕἾ, to draw through Εἰ, F 
respectively two chords AD, AB having a common extremity A and such that DB 
is parallel to EF. The analysis is as follows. Suppose the problem solved, BD 
being parallel to FH. Let BT, the tangent at B, meet HF produced in T. (2 
is not in general the pole of AB, so that ΤᾺ is not generally the tangent at A.) 

Then ZLTBF= 2 BDA, in the alternate segment, 

= 4 AET, by parallels. 
Therefore A, E, B, T are concyclic, and 
EEO AH) ΠΕΣ 
But, the circle ADB and the point F being given, the rectangle AF’. FB is given. 
Also EF is given. 

Hence FT is known. 

Thus, to make the construction, we have only to find the length of FT from 
the data, produce HF to T so that FT has the ascertained length, draw the 
tangent 7'B, and then draw BD parallel to HF. DE, BF will then meet in 4 on 
the circle and will be the chords required. 


BOOK OF LEMMAS. ole 


Thus 
(A 0? + DO’) + (BO? + CO”) = AD? + BC? 
= BE? BC 
= CF”, 


Proposition 12. 


If AB be the diameter of a semicircle, and TP, TQ the 
tangents to ut from any point T, and if AQ, BP be joined 
meeting un R, then TR is perpendicular to AB. 

Let TR produced meet AB in M, and join PA, QB. 


Since the angle APB is right, 


ZPAB+ 2 PBA =(a right angle) 
ΞΘ Ὁ. 


Add to each side the angle RBQ, and 
ZPAB+2ZQBA = (exterior) 2 PRY. 
But ZIPPER ΞΖ ΡΒ. ang 4 TOR=7 OBA, 
in the alternate segments ; 
therefore ZTPR+Z7TQR= 2 PRQ. 
It follows from this that 7P=7TQ=TR. 
[For, if PT’ be produced to O so that TO = TQ, we have 
ZLOO =A 7 D0: 
And, by hypothesis, 2PRQ=2ZTPR+TQR. 
By addition, 2 POQ+ 2 PRQ=2TPR+ OQR. 


314 ARCHIMEDES 


It follows that, in the quadrilateral OPRQ, the opposite 
angles are together equal to two right angles. Therefore a 
circle will go round OPQR, and T is its centre, because 


TP=TO=T7Q. “Therefore TR = TP. | 
Thus ATE = PPR =7:P AM, 
Adding to each the angle PRM, 
ZPAM+2ZPRM=ZTRP+2ZPRM 
= (two right angles). 
Therefore ZAPR+2Z AMR =(two right angles), 
whence Z AMR =(a right angle)*. 


Proposition 13S. 


If a diameter AB of a circle meet any chord CD, not a 
diameter, in 1), and if AM, BN be drawn perpendicular to CD, 
then 

Cy pu 

Let O be the centre of the 
circle, and OH perpendicular to 
CD. Join BM, and produce HO to 
meet BM in K. 


Then CH= Fy. 
And, by parallels, 


since BO — OA, 
ie 
Therefore VG — ΤΙ 
Accordingly CN = DM. 


* TM is of course the polar of the intersection of PQ, AB, as it is the line 
joining the poles of PQ, AB respectively. 

+ This proposition is of course true whether M, N lie on CD or on CD 
produced each way. Pappus proves it for the latter case in his first lemma 
(p. 788) to the second Book of Apollonius’ νεύσεις. 


BOOK OF LEMMAS. 3ol5 


Proposition 14. 


Let ACB be a semicircle on AB as diameter, and let AD, 
BE be equal lengths measured along AB from A, B respectively. 
On AD, BE as diameters describe semicircles on the side towards 
C, and on DE as diameter a semicircle on the opposite side. Let 
the perpendicular to AB through O, the centre of the first semi- 
circle, meet the opposite senucircles in C, F' respectively. 


Then shall the area of the figure bounded by the circumferences 
of all the semicircles (“which Archimedes calls ‘Salinon’”*) be 
equal to the area of the circle on CF as diameter ft. 


By Eucl. τι. 10, since #D is bisected at O and produced 


to A, 
EA’ + AD’ =2(EO’ + 045), 


and CF=0A+0H=EA., 


* For the explanation of this name see note attached to the remarks on the 
Liber Assumptorum in the Introduction, Chapter II. On the grounds there 
given at length I believe σάλινον to be simply a Graecised form of the Latin 
word salinum, ‘salt-cellar.’ 

+ Cantor (Gesch. d. Mathematik, 1. p. 285) compares this proposition 
with Hippocrates’ attempt to square the circle by means of lunes, but 
points out that the object of Archimedes may have been the converse of that 
of Hippocrates. For, whereas Hippocrates wished to find the area of a circle 
from that of other figures of the same sort, Archimedes’ intention was possibly 
to equate the area of figures bounded by different curves to that of a circle 
regarded as already known. 


316 ARCHIMEDES 


Therefore 
AB + DE? =4(EO? + 0A’) =2 (CF? + AD’). 
But circles (and therefore semicircles) are to one another as 
the squares on their radii (or diameters). 
Therefore 
(sum of semicircles on AB, DE’) 


= (circle on CF) + (sum of semicircles on AD, BE). 
Therefore 


(area of ‘salinon’)=(area of circle on CF as diam.). 


Proposition 15. 


Let AB be the diameter of a circle, AC a side of an im- 
scribed regular pentagon, D the middle point of the arc AC. 
Join CD and produce it to meet BA produced in Εἰ; join AC, 
DB meeting in F, and draw FM perpendicular to AB. Then 


EM =(radius of circle)*. 


Let O be the centre of the circle, and join DA, DM, DO, 
CB. 


Now Z ABC =2 (right angle), 
and Z ABD=2 DBC =1 (right angle), 
whence Z AOD =2 (right angle). 


* Pappus gives (p. 418) a proposition almost identical with this among the 
lemmas required for the comparison of the five regular polyhedra. His enunci- 
ation is substantially as follows. If DH be half the side of a pentagon inscribed 
in a circle, while DH is perpendicular to the radius OHA, and if HM be made 
equal to AH, then O4 is divided at M in extreme and mean ratio, OM being the 
greater segment. 


In the course of the proof it is first shown that AD, DM, MO are all equal, 
as in the proposition above. 


Then, the triangles ODA, DAM being similar, 
OA: AD=AD: AM, 
or (since AD = OM) OA: OM=OM : MA. 


BOOK OF LEMMAS. 317 
Further, the triangles FCB, FMB are equal im all respects. 


Therefore, in the triangles DCB, DMB, the sides CB, MB 
being equal and BD common, while the angles CBD, MBD are 
equal, 


ZBCD=2 BMD =§ (right angle). 


But ZBCD+2Z BAD =(two right angles) 
=/BAD+2ZDAE 
= BMD+2zDMA, 

so that ZDAK=z BCD, 
and ZbAp—Z AMD. 

Therefore AD= MD. 


Now, in the triangle DMO, 
Z MOD =2 (right angle), 
Z DMO = ὁ (right angle). 
Therefore 2 ODM=2(right angle) = AOD; 


whence OM = MD. 
Again Z EDA =(supplement of ADC) 
=ZCBA 


= 2 (right angle) 


=2Z ODM. 


318 ARCHIMEDES 
Therefore, in the triangles EDA, ODM, 
ZEDA=2 ODM, 
ZEAD=ZOMD, 
and the sides AD, MD are equal. 
Hence the triangles are equal in all respects, and 
EA = MO. 
Therefore EM = AO. 


Moreover DE = DO; and it follows that, since DE is equal 
to the side of an inscribed hexagon, and DC is the side of an 
inscribed decagon, HC is divided at D in extreme and mean 
ratio [ie. EC: ED=ED: DC]; “and this is proved in the 
book of the Elements.” [πο]. xu. 9, “If the side of the 
hexagon and the side of the decagon inscribed in the same 
circle be put together, the whole straight line is divided in 
extreme and mean ratio, and the greater segment is the 
side of the hexagon.” 


THE CATTLE-PROBLEM. 


ΤΥ is required to find the number of bulls and cows of each 
of four colours, or to find 8 unknown quantities. The first 
part of the problem connects the unknowns by seven simple 
equations; and the second part adds two more conditions to 
which the unknowns must be subject. 


Let W, w be the numbers of white bulls and cows respectively, 


XxX, δὴ ; Fs black = : a 
Vay 7 5 yellow, Z . 
7.2 Ps » dappled _,, τ Υ 
First part. 
(1) ΞΕ τε ease hati ει (α), 
Χ {Ὁ ΖΕ τ ρος (8), 
ZG ee to ie eta cneh ere eae (y), 
(ΠῚ) ᾿ Ξε oe ae (8), 
ΧΦ ΞΞΙ ΙΕ ρα τ (e), 
2 (BEA) τε τ (ζ), 
= EAE CWSU) - (η). 


Second part. 
W +X =a square 


Y+ Z =a triangular number...............(0). 


320 ARCHIMEDES 


[There is an ambiguity in the language which expresses the 
condition (@). Literally the lines mean “ When the white bulls 
joined in number with the black, they stood firm (ἔμπεδον) 
with depth and breadth of equal measurement (ἐσόμετροι εἰς 
βάθος εἰς εὖρός Te); and the plains of Thrinakia, far-stretching 
all ways, were filled with their multitude” (reading, with 
Krumbiegel, πλήθους instead of πλίνθου). Considering that, if 
the bulls were packed together so as to form a square figure, 
the number of them need not be a square number, since a buil 
is longer than it is broad, it is clear that one possible interpre- 
tation would be to take the ‘square’ to be a square figure, and 
to understand condition (@) to be simply 

W + X =a rectangle (i.e. a product of two factors). 
The problem may therefore be stated in two forms: 
(1) the simpler one in which, for the condition (@), there is 
substituted the mere requirement that 
W+X =a product of two whole numbers ; 
(2) the complete problem in which all the conditions have to 
be satisfied including the requirement (@) that 
W + X =a square number. 

The simpler problem was solved by Jul. Fr. Wurm and may 

be called 


Wurm/’s Problem. 


The solution of this is given (together with a discussion of 
the complete problem) by Amthor in the Zeitschrift fiir Math. 
u. Physik (Hist. litt. Abtheilung), XXv. (1880), p. 156 sqq. 


Multiply (a) by 336, (8) by 280, (y) by 126, and add; thus 
207 W=7142Y, or 3°. 11 W 2.7 5 
Then from (γ) and (8) we obtain 
8917 =1580Y, or 3'.11Z7=2.5.79Y......(8), 
and 90X = ΠΟῪ, or 3? 1LX ]2 δ... τ 
Again, if we multiply (δ) by 4800, (e) by 2800, (ζ) by 1260, 
(η) by 462, and add, we obtain 
4657w = 2800X + 12607 + 462Y +148 W ; 


THE CATTLE-PROBLEM. 3 yall 


and, by means of the values in (a’), (8), (y’), we derive 


297. 4657w = 24021207, 


or a tb Aootm— 2.5.7. 23.873 τ... {ὃ} 
Hence, by means of (7), (ζ), (ε), we have 

3°.11.4657y = 13.46489Y...... Bena (e’), 

oF aoe OF 75.0 0 ai cese ct (&) 

and o-oo 210 SUS9OTY cs. cccace (7’). 


And, since all the unknowns must be whole numbers, we see 
from the equations (a’), (8’), ... (7’) that Y must be divisible by 
3°.11. 4657, 1.6. we may put 


Y = 3+.11.4657n = 4149387n. 


Therefore the equations (a’), (8’),...(n’) give the following values 
for all the unknowns in terms of n, viz. 


W =2.3.7.53.4657n =10366482n 


X = 2.37.89. 4657” = 7460514n 

V3. Ut Ahan = 4149387n 
Ζ--.25.ὅ.19. 465 7H = 7358060n\ (A). 
w = 2°.3.5.7.23.3738n = 7206360n/ 
2=2.37.17.15991n = 4893246n 

y = 9.13. 46489n = 5439213n 


2— 2, πὴ lie voln— 95 1052 05 


If now n = 1, the numbers are the smallest which will satisfy 
the seven equations (a), (8),...(7); and we have next to find 
such an integral value for n that the equation (2) will be 
satisfied also. [The modified equation (@) requiring that W +X 
must be a product of two factors is then simultaneously 
satisfied. | 


Equation (4) requires that 
γι Foie) 
9 ") 


where q is some positive integer. 


HA. 21 


322 


ARCHIMEDES 


Putting for Y, Z their values as above ascertained, we have 


oe =(3*.11 427.5. 79). 4657n 


= 2471. 4657n 
= 7.353 .4657n. 


Now q is either even or odd, so that either g= 2s, or 
qg=2s—1, and the equation becomes 


s(2s + 1)=7.353.4657n. 


As n need not be a prime number, we suppose Ἢ = w. v, where 
wu is the factor in n which divides s without a remainder and ὃ 
the factor which divides 2s +1 without a remainder; we then 
have the following sixteen alternative pairs of simultaneous 


equations : 


(1) 
(2) 
(3) 
(4) 
(5) 
(6) 
(7) 
(8) 


ΒΞ U, 
— Tu, 
a 353 u, 
s= 4657 u, 
— 7.35384, 
— 7.4657, 


s= 353.4657u, 
s=7.353.4657u, 


2s8+1=7.353. 4657, 


2s+1= 
28 41= 
2s84+1= 
23 4 
25 1-Ξ 
%+1= 
23+ l= 


353. 46570, 
7.46572, 
7.3530, 
46579, 
3532, 

Tv, 


υ. 


In order to find the least value of n which satisfies all the 
conditions of the problem, we have to choose from the various 
positive integral solutions of these pairs of equations that 
particular one which gives the smallest value for the product 


Uv Or Nn. 


If we solve the various pairs and compare the results, we 
find that it is the pair of equations 


s=7u, 2s—1=853. 4657, 


which leads to the solution we want; this solution is then 


so that 


u = 117423, 


ise 


n = uw = 117423 = 3°. 4349, 


THE CATTLE-PROBLEM. ἘΣ A, 


whence it follows that 


S70) — 021061, 


and q = 2s — 1 = 1643921. 
Thus VY 4+ Z= 2471. 4657n 
= 2411. 4657 . 117423 
= 1351238949081 
_ 1643921 . 1643922 
we 


which is a triangular number, as required. 


The number in equation (0) which has to be the product of 
two integers is now 
W+X =2.3.(7.5384 3.89). 4657n 
= 2?.3.11.29.4657n 
= 2’.3.11.29. 4657. 117423 
= 2? 3*.11.29. 4657 . 4349 
= (2°. 3°. 4349). (11. 29 . 4657) 
= 1409076 . 1485583, 


which is a rectangular number with nearly equal factors. 


The solution is then as follows (substituting for n its value 
117423): 

W = 1217263415886 

X = 876035935422 

= 487233469701 

Z= 864005479380 

w= 846192410280 

x= 574579625058 

y = 638688708099 

z= 412838151860 

and the sum = 5916837175686 


324 ARCHIMEDES 
The complete problem. 


In this case the seven original equations (a), (8),...(7) have 
to be satisfied, and the following further conditions must hold, 


W + X =a square number = p’, say, 


q(qt1) 
2 


Using the values found above (A), we have in the first place 
p=2.3.(7.53+3.89).4657n 
=2?, 3.11.29. 4657n, 
and this equation will be satisfied if 
n=3.11.29.4657& = 4450149 ξ", 


where & is any integer. 


Y + Z=a triangular number = , Say. 


Thus the first 8 equations (a), (8),...(m), (0) are satisfied by 
the following values : 


W =2.37.7.11.29.53. 4657". & = 46200808287018 . E* 
BGs eal OAS 0. 7: Σ- = 33249638308986 . ξ" 
Y=3'.11°. 29.4657’. & = 18492776362863. &° 
La 35 ὙΠ 979 100 γ΄...’ = 32793026546940. &° 


w= 2 .3?.5.7.11. 23.29.3738. 4657 . ξ΄ = 32116937723640. &* 
g=2.3°.11.17.29.15991 .4657.2" =21807969217254. &* 
y=93*.11.18.29. 46489 . 4657 . ξ" = 24241207098537 . &” 
g=2°,37.5.7.11°.29. 761.4657 .é" =15669127269180R 
It remains to determine ἕ so that equation (4) may be 
satisfied, 1.e. so that 
τὰ χεύτυ, 
2 
Substituting the ascertained values of Y, Z, we have 


nes = 51285802909803 . ξ᾽ 


=3.7.11.29.353.4657?. £ 


THE CATTLE-PROBLEM. 325 
Multiply by 8, and put 
Qqt1l=t, 2.4657.€=4u, 
and we have the “ Pellian” equation 
πο 7. VE 29. 853-0, 
that is, & — 4729494 uv? = 1. 


Of the solutions of this equation the smallest has to be 
chosen for which w is divisible by 2 . 4657. 


When this is done, 


and is a whole number ; 


U 
δ = 94657 


whence, by substitution of the value of € so found in the last 
system of equations, we should arrive at the solution of the 
complete problem. 


It would require too much space to enter on the solution of 
the “ Pellian” equation 
t — 4729494 wv? = 1, 


and the curious reader is referred to Amthor’s paper itself. 
Suffice it to say that he develops V4729494 in the form of a 
continued fraction as far as the period which occurs after 91 
convergents, and, after an arduous piece of work, arrives at the 
conclusion that 


W =1598 <2065H>, 


where <206541 > represents the fact that there are 206541 more 
digits to follow, and that, with the same notation, 


the whole number of cattle = 7766 <206541>, 


One may well be excused for doubting whether Archimedes 
solved the complete problem, having regard to the enormous 


326 ARCHIMEDES 


size of the numbers and the great difficulties inherent in the 
work. By way of giving an idea of the space which would be 
required for merely writing down the results when obtained, 
Amthor remarks that the large seven-figured logarithmic tables 
contain on one page 50 lines with 50 figures or so in each, say 
altogether 2500 figures; therefore one of the eight unknown 
quantities would, when found, occupy 824 such pages, and to 
write down all the eight numbers would require a volume of 
660 pages !] 


CAMBRIDGE: PRINTED BY J. AND C. F. CLAY, AT THE UNIVERSITY PRESS. 


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